Samacheer Kalvi Class 12 Maths Solutions Chapter 6 Applications of Vector Algebra Exercise 6.1

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1

Question 1. Prove by vector method that if a line is drawn from the centre of a circle to the midpoint of a chord; then the line is perpendicular to the chord.
Answer: Let O be the center of the circle. Let AB be a chord of this circle. Let P be the midpoint of the chord AB. This means \( AP = PB \). We need to show that the line segment OP (drawn from the center to the midpoint) is perpendicular to the chord AB. We know that vectors from the center to any point on the circle have the same magnitude (radius).
\[ \overline{OP} = \frac { \overline{OA} + \overline{OB} }{ 2 } \] Next, we calculate the dot product of \( \overline{OP} \) and \( \overline{AB} \):
\[ \overline{OP} \cdot \overline{AB} = \left( \frac { \overline{OA} + \overline{OB} }{ 2 } \right) \cdot (\overline{OB} - \overline{OA}) \] This is similar to the algebraic identity \( (x+y)(x-y) = x^2 - y^2 \).
\[ = \frac { 1 }{ 2 } (|\overline{OB}|^2 - |\overline{OA}|^2) \] Since \( \overline{OA} \) and \( \overline{OB} \) are both radii of the same circle, their magnitudes are equal. The radius is the distance from the center to any point on the circle.
\[ |\overline{OA}| = |\overline{OB}| = \text{Radius} \] So, \( |\overline{OB}|^2 - |\overline{OA}|^2 = 0 \).
\[ \overline{OP} \cdot \overline{AB} = 0 \] Since the dot product is zero, the vectors \( \overline{OP} \) and \( \overline{AB} \) are perpendicular to each other. This proves that the line from the center to the midpoint of a chord is perpendicular to the chord.
In simple words: When you draw a line from the middle of a circle to the exact center of a chord (a line across the circle), that line will always meet the chord at a perfect right angle (90 degrees).

🎯 Exam Tip: Remember that the dot product of two non-zero vectors is zero if and only if they are perpendicular. This is a key concept in vector proofs involving perpendicularity.

Question 2. Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base.
Answer: Let's consider an isosceles triangle \( \triangle ABC \) where \( AB = AC \). Let AD be the median to the base BC, meaning D is the midpoint of BC. We need to prove that AD is perpendicular to BC. An isosceles triangle has two sides of equal length, which is key to this proof.
We can write the position vector of D as the average of the position vectors of B and C:
\[ \overline{AD} = \frac { \overline{AB} + \overline{AC} }{ 2 } \] The base vector BC can be expressed as:
\[ \overline{BC} = \overline{BA} + \overline{AC} \] Now, let's calculate the dot product of \( \overline{AD} \) and \( \overline{BC} \). To make the calculation easier, we can express \( \overline{DA} \) and \( \overline{DB} \):
\[ \overline{DA} = - \overline{AD} \] Since D is the midpoint of BC, \( \overline{DB} = - \frac { 1 }{ 2 } \overline{CB} \).
\[ \overline{DA} \cdot \overline{DB} = \left( - \frac { \overline{AB} + \overline{AC} }{ 2 } \right) \cdot \left( - \frac { 1 }{ 2 } \overline{CB} \right) \] \[ = \frac { 1 }{ 4 } (\overline{AB} + \overline{AC}) \cdot (\overline{CB}) \] We know \( \overline{CB} = \overline{AB} - \overline{AC} \). Substituting this:
\[ = \frac { 1 }{ 4 } (\overline{AB} + \overline{AC}) \cdot (\overline{AB} - \overline{AC}) \] Using the difference of squares formula \( (x+y)(x-y) = x^2 - y^2 \):
\[ = \frac { 1 }{ 4 } (|\overline{AB}|^2 - |\overline{AC}|^2) \] Since \( \triangle ABC \) is an isosceles triangle with \( AB = AC \), their magnitudes are equal:
\[ |\overline{AB}| = |\overline{AC}| \] Therefore, \( |\overline{AB}|^2 - |\overline{AC}|^2 = 0 \).
\[ \overline{DA} \cdot \overline{DB} = 0 \] Since the dot product of \( \overline{DA} \) and \( \overline{DB} \) is zero, the median AD is perpendicular to the base BC. This property is unique to isosceles and equilateral triangles.
In simple words: In a triangle where two sides are the same length (isosceles triangle), if you draw a line from the top corner to the middle of the bottom side, that line will always be at a right angle to the bottom side.

🎯 Exam Tip: When proving perpendicularity using vectors, the main goal is to show that the dot product of the two relevant vectors is zero.

Question 3. Prove by vector method that an angle in a semi-circle is a right angle.
Answer: Let's consider a circle with center O and diameter AB. Let P be any point on the semi-circle. We want to prove that the angle \( \angle APB \) is 90 degrees. This is a classic geometry theorem that is elegantly proven with vectors.
We know that OA, OB, and OP are all radii of the circle, so their magnitudes are equal:
\[ |\overline{OA}| = |\overline{OB}| = |\overline{OP}| = \text{radius} \] Since O is the midpoint of the diameter AB, we have \( \overline{OB} = - \overline{OA} \).
Now, let's find the vectors \( \overline{PA} \) and \( \overline{PB} \):
\[ \overline{PA} = \overline{PO} + \overline{OA} \] \[ \overline{PB} = \overline{PO} + \overline{OB} \] Substitute \( \overline{OB} = - \overline{OA} \) into the expression for \( \overline{PB} \):
\[ \overline{PB} = \overline{PO} - \overline{OA} \] Now, let's calculate the dot product of \( \overline{PA} \) and \( \overline{PB} \):
\[ \overline{PA} \cdot \overline{PB} = (\overline{PO} + \overline{OA}) \cdot (\overline{PO} - \overline{OA}) \] Using the difference of squares formula \( (x+y)(x-y) = x^2 - y^2 \):
\[ = |\overline{PO}|^2 - |\overline{OA}|^2 \] Since \( |\overline{PO}| \) and \( |\overline{OA}| \) are both equal to the radius, their squares are also equal.
\[ = (\text{radius})^2 - (\text{radius})^2 = 0 \] Therefore, \( \overline{PA} \cdot \overline{PB} = 0 \).
This means that the vectors \( \overline{PA} \) and \( \overline{PB} \) are perpendicular. So, the angle \( \angle APB \) is 90 degrees. This property holds true for any point P on the semi-circle.
In simple words: If you draw a triangle inside a circle, and one side of the triangle is the circle's diameter, then the angle opposite to that diameter will always be a right angle (90 degrees).

🎯 Exam Tip: Clearly defining position vectors and using the property of radii being equal are crucial steps in this proof. Remember to use the difference of squares identity for dot products.

Question 4. Prove by vector method that the diagonals of a rhombus bisect each other at right angles.
Answer: Let ABCD be a rhombus. In a rhombus, all four sides are equal in length. This is the defining characteristic that will help us in the vector proof. Let \( \overline{AB} = \overline{a} \) and \( \overline{AD} = \overline{b} \). Since all sides are equal, \( |\overline{a}| = |\overline{b}| \).
The diagonals of the rhombus are \( \overline{AC} \) and \( \overline{BD} \).
Using vector addition for \( \overline{AC} \):
\[ \overline{AC} = \overline{AB} + \overline{BC} \] Since ABCD is a rhombus, \( \overline{BC} = \overline{AD} = \overline{b} \).
\[ \overline{AC} = \overline{a} + \overline{b} \] Using vector addition for \( \overline{BD} \):
\[ \overline{BD} = \overline{BA} + \overline{AD} \] Since \( \overline{BA} = - \overline{AB} = - \overline{a} \).
\[ \overline{BD} = - \overline{a} + \overline{b} = \overline{b} - \overline{a} \] Now, let's find the dot product of the diagonals \( \overline{AC} \) and \( \overline{BD} \):
\[ \overline{AC} \cdot \overline{BD} = (\overline{a} + \overline{b}) \cdot (\overline{b} - \overline{a}) \] We can rearrange this as:
\[ = (\overline{b} + \overline{a}) \cdot (\overline{b} - \overline{a}) \] Using the difference of squares identity \( (x+y)(x-y) = x^2 - y^2 \):
\[ = |\overline{b}|^2 - |\overline{a}|^2 \] Since it's a rhombus, \( |\overline{a}| = |\overline{b}| \). Therefore, \( |\overline{b}|^2 - |\overline{a}|^2 = 0 \).
\[ \overline{AC} \cdot \overline{BD} = 0 \] Since the dot product is zero, the diagonals \( \overline{AC} \) and \( \overline{BD} \) are perpendicular to each other. Additionally, it is a known property that the diagonals of any parallelogram (and thus any rhombus) bisect each other. So, this proof shows they bisect each other at right angles.
In simple words: In a rhombus (a shape with four equal sides), the two lines drawn from corner to opposite corner (diagonals) always cut each other into two equal halves. Also, these lines meet at a perfect right angle.

🎯 Exam Tip: The key to proving the perpendicularity of diagonals in a rhombus is recognizing that all sides are equal, which makes the magnitudes of the side vectors equal, leading to the cancellation in the dot product.

Question 5. Using vector method, prove that if the diagonals of a parallelogram are equal, then it is a rectangle.
Answer: Let ABCD be a parallelogram. In a parallelogram, opposite sides are equal and parallel. If its diagonals are equal in length, we need to prove it is a rectangle. A rectangle is a parallelogram with all angles equal to 90 degrees.
Let \( \overline{AB} = \overline{a} \) and \( \overline{AD} = \overline{b} \).
Then \( \overline{BC} = \overline{AD} = \overline{b} \) and \( \overline{CD} = \overline{BA} = - \overline{a} \).
The diagonals are \( \overline{AC} \) and \( \overline{BD} \).
Using vector addition:
\[ \overline{AC} = \overline{AB} + \overline{BC} = \overline{a} + \overline{b} \] \[ \overline{BD} = \overline{BA} + \overline{AD} = - \overline{a} + \overline{b} = \overline{b} - \overline{a} \] The problem states that the diagonals are equal, so their magnitudes are equal:
\[ |\overline{AC}| = |\overline{BD}| \] Squaring both sides:
\[ |\overline{AC}|^2 = |\overline{BD}|^2 \] \[ |\overline{a} + \overline{b}|^2 = |\overline{b} - \overline{a}|^2 \] Expanding the magnitudes:
\[ (\overline{a} + \overline{b}) \cdot (\overline{a} + \overline{b}) = (\overline{b} - \overline{a}) \cdot (\overline{b} - \overline{a}) \] \[ |\overline{a}|^2 + 2\overline{a} \cdot \overline{b} + |\overline{b}|^2 = |\overline{b}|^2 - 2\overline{b} \cdot \overline{a} + |\overline{a}|^2 \] Subtract \( |\overline{a}|^2 \) and \( |\overline{b}|^2 \) from both sides:
\[ 2\overline{a} \cdot \overline{b} = -2\overline{a} \cdot \overline{b} \] (Since dot product is commutative, \( \overline{b} \cdot \overline{a} = \overline{a} \cdot \overline{b} \)).
\[ 4\overline{a} \cdot \overline{b} = 0 \] \[ \overline{a} \cdot \overline{b} = 0 \] Since the dot product of \( \overline{a} \) (which is \( \overline{AB} \)) and \( \overline{b} \) (which is \( \overline{AD} \)) is zero, these two adjacent sides are perpendicular to each other. This means the angle \( \angle A \) is 90 degrees. Because it's a parallelogram, all angles must be 90 degrees, making it a rectangle.
In simple words: If a four-sided shape (parallelogram) has diagonals (lines connecting opposite corners) that are the same length, then that shape must be a rectangle. This means all its corners are right angles.

🎯 Exam Tip: The key insight here is to use the property that if the dot product of two adjacent side vectors is zero, the angle between them is 90 degrees, thus proving it's a rectangle.

Question 6. Prove by vector method that the area of the quadrilateral ABCD having diagonals AC and BD is \( \frac { 1 }{ 2 } |\overline { AC } \times \overline { BD }| \).
Answer: Let ABCD be a quadrilateral with diagonals \( \overline{AC} \) and \( \overline{BD} \). We want to prove its area using vectors. The area of any quadrilateral can be found using the cross product of its diagonals. This method simplifies finding the area when only the diagonals are known.
The vector area of a quadrilateral ABCD can be expressed as the sum of the vector areas of two triangles formed by one diagonal:
\[ \text{Vector Area of Quadrilateral ABCD} = \text{Vector Area of } \triangle ABC + \text{Vector Area of } \triangle ACD \] The vector area of a triangle formed by two vectors \( \overline{X} \) and \( \overline{Y} \) is \( \frac { 1 }{ 2 } (\overline{X} \times \overline{Y}) \).
\[ = \frac { 1 }{ 2 } (\overline{AB} \times \overline{AC}) + \frac { 1 }{ 2 } (\overline{AC} \times \overline{AD}) \] Using the property \( \overline{AB} \times \overline{AC} = - (\overline{AC} \times \overline{AB}) \), we can factor out \( \frac { 1 }{ 2 } \overline{AC} \):
\[ = \frac { 1 }{ 2 } (-\overline{AC} \times \overline{AB}) + \frac { 1 }{ 2 } (\overline{AC} \times \overline{AD}) \] \[ = \frac { 1 }{ 2 } \overline{AC} \times (-\overline{AB} + \overline{AD}) \] We know that \( -\overline{AB} + \overline{AD} = \overline{BA} + \overline{AD} \). From the triangle BAD, the vector sum \( \overline{BA} + \overline{AD} \) gives the vector \( \overline{BD} \).
\[ = \frac { 1 }{ 2 } \overline{AC} \times \overline{BD} \] Therefore, the vector area of the quadrilateral ABCD is \( \frac { 1 }{ 2 } \overline{AC} \times \overline{BD} \). To find the actual scalar area, we take the magnitude of this vector:
\[ \text{Area of Quadrilateral ABCD} = \frac { 1 }{ 2 } |\overline{AC} \times \overline{BD}| \] This formula holds for any convex quadrilateral.
In simple words: To find the space covered by a four-sided shape using vectors, you can multiply its two diagonal lines using the cross product method. Then, take half of the size (magnitude) of the result. This gives you the area.

🎯 Exam Tip: Remember the formula for the vector area of a triangle \( \frac{1}{2} (\overline{a} \times \overline{b}) \). The property of vector cross product \( \overline{A} \times \overline{B} = -(\overline{B} \times \overline{A}) \) is crucial for factorization in this proof.

Question 7. Prove by vector method that the parallelograms on the same base and between the same parallels are equal in area.
Answer: Let's consider two parallelograms, ABCD and ABB'A', that share the same base AB and are between the same parallel lines. This means that the height of both parallelograms with respect to base AB will be the same. The area of a parallelogram is given by the magnitude of the cross product of its adjacent sides.
Let the base vector be \( \overline{AB} = \overline{a} \).
For parallelogram ABCD, let \( \overline{AD} = \overline{b} \).
The vector area of parallelogram ABCD is \( \overline{AB} \times \overline{AD} = \overline{a} \times \overline{b} \).
Now, consider parallelogram ABB'A'. It shares the same base AB. Its side \( \overline{AA'} \) connects points on the parallel line. Since A' is on the line parallel to AB through D, \( \overline{AA'} \) can be written as \( m\overline{a} + \overline{b} \) for some scalar m, because \( \overline{AD} = \overline{b} \) and \( \overline{A'B} \) is parallel to \( \overline{AB} \).
So, \( \overline{AA'} = m\overline{AB} + \overline{AD} = m\overline{a} + \overline{b} \).
The vector area of parallelogram ABB'A' is \( \overline{AB} \times \overline{AA'} \).
\[ = \overline{a} \times (m\overline{a} + \overline{b}) \] Using the distributive property of the cross product:
\[ = m(\overline{a} \times \overline{a}) + (\overline{a} \times \overline{b}) \] We know that the cross product of a vector with itself is zero (since the angle between them is 0 degrees, and \( \sin 0 = 0 \)).
\[ = m(0) + (\overline{a} \times \overline{b}) \] \[ = \overline{a} \times \overline{b} \] This shows that the vector area of parallelogram ABB'A' is the same as the vector area of parallelogram ABCD. Since their vector areas are equal, their magnitudes (which represent the scalar areas) are also equal. This proves that parallelograms on the same base and between the same parallels have equal areas.
In simple words: If you have two parallelogram shapes that both sit on the same bottom line and their top lines are also parallel to the bottom line, then both shapes will cover the exact same amount of space (have the same area).

🎯 Exam Tip: The critical step here is expressing the side vector \( \overline{AA'} \) of the second parallelogram in terms of the base vector \( \overline{a} \) and the height-defining vector \( \overline{b} \), and then utilizing the property \( \overline{a} \times \overline{a} = 0 \).

Question 8. If G is the centroid of a \( \triangle ABC \), prove that (area of \( \triangle GAB \)) = (area of \( \triangle GBC \)) = (area of \( \triangle GCA \)) \( = \frac { 1 }{ 3 } \) (area of \( \triangle ABC \)).
Answer: Let \( \triangle ABC \) be a triangle, and let G be its centroid. The centroid is the point where the medians of the triangle intersect. A key property of medians is that they divide the triangle into two triangles of equal area. This geometric fact is the foundation of this vector proof.
Let AD be a median of \( \triangle ABC \) (so D is the midpoint of BC). We know that the median divides the triangle into two equal areas:
\[ \text{area}(\triangle ABD) = \text{area}(\triangle ACD) \quad ..........(1) \] Now, consider \( \triangle GBC \). GD is a median in \( \triangle GBC \) (since G is the centroid and D is the midpoint of BC). Therefore, GD divides \( \triangle GBC \) into two equal areas:
\[ \text{area}(\triangle GBD) = \text{area}(\triangle GCD) \quad ..........(2) \] Subtracting equation (2) from equation (1):
\[ \text{area}(\triangle ABD) - \text{area}(\triangle GBD) = \text{area}(\triangle ACD) - \text{area}(\triangle GCD) \] This subtraction results in:
\[ \text{area}(\triangle AGB) = \text{area}(\triangle AGC) \quad ..........(3) \] Similarly, if we consider median BE (where E is the midpoint of AC), we can prove that:
\[ \text{area}(\triangle AGB) = \text{area}(\triangle BGC) \quad ..........(4) \] From (3) and (4), we get:
\[ \text{area}(\triangle AGB) = \text{area}(\triangle AGC) = \text{area}(\triangle BGC) \quad ..........(5) \] Now, the sum of the areas of the three smaller triangles formed by the centroid is equal to the area of the main triangle:
\[ \text{area}(\triangle AGB) + \text{area}(\triangle AGC) + \text{area}(\triangle BGC) = \text{area}(\triangle ABC) \] Using equation (5), we can replace \( \text{area}(\triangle AGC) \) and \( \text{area}(\triangle BGC) \) with \( \text{area}(\triangle AGB) \):
\[ \text{area}(\triangle AGB) + \text{area}(\triangle AGB) + \text{area}(\triangle AGB) = \text{area}(\triangle ABC) \] \[ 3 \cdot \text{area}(\triangle AGB) = \text{area}(\triangle ABC) \] So, for each smaller triangle:
\[ \text{area}(\triangle AGB) = \frac { 1 }{ 3 } \text{area}(\triangle ABC) \quad ..........(6) \] From (5) and (6), we conclude that all three triangles formed by the centroid have equal areas, and each area is one-third of the total triangle's area.
\[ \text{area}(\triangle AGB) = \text{area}(\triangle BGC) = \text{area}(\triangle GCA) = \frac { 1 }{ 3 } \text{area}(\triangle ABC) \]
In simple words: The centroid is like the balance point of a triangle. If you draw lines from this center point to each corner, you will divide the big triangle into three smaller triangles. All these three smaller triangles will have the exact same amount of space (area), and each will be one-third of the whole triangle's area.

🎯 Exam Tip: The core idea is that a median divides a triangle into two equal areas. Applying this property repeatedly to the main triangle and the smaller triangles formed by the centroid will lead to the desired result.

Question 9. Prove that \( \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \).
Answer: To prove the cosine difference formula using vectors, let's consider two unit vectors, \( \overline{a} \) and \( \overline{b} \), originating from the origin O. Unit vectors have a magnitude of 1, which simplifies calculations significantly. Let these vectors make angles \( \alpha \) and \( \beta \) respectively with the positive x-axis.
The coordinates of point A (endpoint of \( \overline{a} \)) would be \( (\cos \alpha, \sin \alpha) \).
So, the unit vector \( \overline{a} \) can be written as: \( \overline{a} = \cos \alpha \hat{i} + \sin \alpha \hat{j} \).
Similarly, the coordinates of point B (endpoint of \( \overline{b} \)) would be \( (\cos \beta, \sin \beta) \).
So, the unit vector \( \overline{b} \) can be written as: \( \overline{b} = \cos \beta \hat{i} + \sin \beta \hat{j} \).
The angle between these two vectors, \( \overline{a} \) and \( \overline{b} \), is \( \alpha - \beta \).
The dot product of two vectors is also given by the formula \( \overline{a} \cdot \overline{b} = |\overline{a}| |\overline{b}| \cos \theta \), where \( \theta \) is the angle between them.
Since \( \overline{a} \) and \( \overline{b} \) are unit vectors, \( |\overline{a}| = 1 \) and \( |\overline{b}| = 1 \). The angle between them is \( \alpha - \beta \).
So, \( \overline{a} \cdot \overline{b} = (1)(1) \cos(\alpha - \beta) = \cos(\alpha - \beta) \quad ..........(3) \)
Now, let's calculate the dot product using their component forms:
\[ \overline{a} \cdot \overline{b} = (\cos \alpha \hat{i} + \sin \alpha \hat{j}) \cdot (\cos \beta \hat{i} + \sin \beta \hat{j}) \] \[ = \cos \alpha \cos \beta (\hat{i} \cdot \hat{i}) + \cos \alpha \sin \beta (\hat{i} \cdot \hat{j}) + \sin \alpha \cos \beta (\hat{j} \cdot \hat{i}) + \sin \alpha \sin \beta (\hat{j} \cdot \hat{j}) \] Since \( \hat{i} \cdot \hat{i} = 1 \), \( \hat{j} \cdot \hat{j} = 1 \), and \( \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{i} = 0 \):
\[ \overline{a} \cdot \overline{b} = \cos \alpha \cos \beta + \sin \alpha \sin \beta \quad ..........(4) \] Comparing equations (3) and (4):
\[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \] This proves the cosine difference identity.
In simple words: This formula helps you find the cosine of the difference between two angles. It says you can get the answer by multiplying the cosines of the two angles, then adding that to the product of their sines. It's a key rule in trigonometry.

🎯 Exam Tip: The use of unit vectors makes this proof elegant. Remember that the dot product \( \overline{a} \cdot \overline{b} \) can be calculated in two ways: using magnitudes and the angle between them, and using their component forms. Equating these two expressions leads to the identity.

Question 10. Prove by vector method that \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \).
Answer: To prove the sine addition formula using vectors, we will again use unit vectors and their components. This time, the cross product will be useful. Let \( \overline{a} \) and \( \overline{b} \) be unit vectors making angles \( \alpha \) and \( \beta \) with the positive x-axis, respectively. A unit vector simply means its length is 1, which helps simplify calculations.
Vector \( \overline{a} \) can be written as: \( \overline{a} = \cos \alpha \hat{i} + \sin \alpha \hat{j} \).
Vector \( \overline{b} \) can be written as: \( \overline{b} = \cos \beta \hat{i} + \sin \beta \hat{j} \).
The angle between \( \overline{a} \) and \( \overline{b} \) is \( \alpha + \beta \), but to use cross product effectively for sine, we often need to rotate one vector. Let's use a modified vector for \( \overline{b} \) where its angle is \( -\beta \) or use the angle for the cross product as \( \alpha - (-\beta) = \alpha + \beta \) directly if the direction is chosen carefully. Let's consider \( \overline{a} \) and a vector \( \overline{b'} \) that makes an angle \( -\beta \) with the x-axis. Then \( \overline{b'} = \cos(-\beta)\hat{i} + \sin(-\beta)\hat{j} = \cos\beta\hat{i} - \sin\beta\hat{j} \). The angle between \( \overline{a} \) and \( \overline{b'} \) is \( \alpha - (-\beta) = \alpha + \beta \). The cross product of \( \overline{a} \) and \( \overline{b'} \) is given by \( \overline{a} \times \overline{b'} = |\overline{a}||\overline{b'}| \sin(\alpha + \beta) \hat{k} \). Since they are unit vectors, \( |\overline{a}| = |\overline{b'}| = 1 \).
\[ \overline{a} \times \overline{b'} = \sin(\alpha + \beta) \hat{k} \quad ..........(1) \] Now, let's compute the cross product using their components:
\[ \overline{a} \times \overline{b'} = (\cos \alpha \hat{i} + \sin \alpha \hat{j}) \times (\cos \beta \hat{i} - \sin \beta \hat{j}) \] Using the distributive property of the cross product and knowing \( \hat{i} \times \hat{i} = 0 \), \( \hat{j} \times \hat{j} = 0 \), \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{i} = -\hat{k} \):
\[ = (\cos \alpha)(-\sin \beta) (\hat{i} \times \hat{j}) + (\sin \alpha)(\cos \beta) (\hat{j} \times \hat{i}) \] \[ = -\cos \alpha \sin \beta \hat{k} + \sin \alpha \cos \beta (-\hat{k}) \] \[ = (\sin \alpha \cos \beta + \cos \alpha \sin \beta) \hat{k} \quad ..........(2) \] Comparing equations (1) and (2):
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] This proves the sine addition identity.
In simple words: This formula helps you find the sine of the sum of two angles. It says you multiply the sine of the first angle by the cosine of the second, then add that to the product of the cosine of the first angle and the sine of the second.

🎯 Exam Tip: For sine identities, the cross product is often more direct than the dot product. Be careful with the sign conventions of \( \hat{i} \times \hat{j} \) and \( \hat{j} \times \hat{i} \), and ensure the angle used in the cross product matches the one in the identity.

Question 11. A particle acted on by constant forces \( \overline { F_1 } = 8\hat { i } + 2\hat { j } – 6\hat { k } \) and \( \overline { F_2 } = 6\hat { i } + 2\hat { j } – 2\hat { k } \) is displaced from the point (1, 2, 3) to the point (5, 4, 1). Find the total work done by the forces.
Answer: To find the total work done, we first need to find the resultant force and the displacement vector. Work done by a constant force is the dot product of the force and displacement. This calculation is a straightforward application of vector operations in physics.
The position vector of the initial point A(1, 2, 3) is \( \overline{OA} = \hat{i} + 2\hat{j} + 3\hat{k} \).
The position vector of the final point B(5, 4, 1) is \( \overline{OB} = 5\hat{i} + 4\hat{j} + \hat{k} \).
The displacement vector \( \overline{d} \) is \( \overline{AB} = \overline{OB} - \overline{OA} \):
\[ \overline{d} = (5-1)\hat{i} + (4-2)\hat{j} + (1-3)\hat{k} \] \[ \overline{d} = 4\hat{i} + 2\hat{j} - 2\hat{k} \] Next, we find the resultant force \( \overline{F} \) by adding the two given forces:
\[ \overline{F} = \overline{F_1} + \overline{F_2} \] \[ \overline{F} = (8\hat{i} + 2\hat{j} - 6\hat{k}) + (6\hat{i} + 2\hat{j} - 2\hat{k}) \] \[ \overline{F} = (8+6)\hat{i} + (2+2)\hat{j} + (-6-2)\hat{k} \] \[ \overline{F} = 14\hat{i} + 4\hat{j} - 8\hat{k} \] Now, the work done is the dot product of the resultant force and the displacement vector:
\[ \text{Work done} = \overline{F} \cdot \overline{d} \] \[ = (14\hat{i} + 4\hat{j} - 8\hat{k}) \cdot (4\hat{i} + 2\hat{j} - 2\hat{k}) \] Multiply the corresponding components and sum them up:
\[ = (14)(4) + (4)(2) + (-8)(-2) \] \[ = 56 + 8 + 16 \] \[ = 80 \text{ units} \] The total work done by the forces is 80 units.
In simple words: First, add up all the push/pull forces acting on the particle to get one total force. Then, figure out how far and in what direction the particle moved. Finally, multiply the total force by the distance moved in that force's direction to find the total work done.

🎯 Exam Tip: Remember that work done is a scalar quantity (just a number), so you use the dot product of force and displacement. Always find the resultant force first if multiple forces are acting, and ensure correct calculation of the displacement vector.

Question 12. Forces of magnitudes \( 5\sqrt{2} \) and \( 10\sqrt{2} \) units acting in the directions \( 3\hat { i } + 4\hat { j } + 5\hat { k } \) and \( 10\hat { i } + 6\hat { j } – 8\hat { k } \), respectively, act on a particle which is displaced from the point with position vector \( 4\hat { i } – 3\hat { j } – 2\hat { k } \) to the point with position vector \( 6\hat { i } – \hat { j } – 3\hat { k } \). Find the work done by the forces.
Answer: To find the work done, we first need to determine the actual force vectors and the displacement vector. A force vector is found by multiplying its magnitude by its direction unit vector. Work done is calculated by the dot product of the resultant force and displacement. This problem involves normalizing direction vectors before scaling them by magnitudes.
First, let's find the unit vectors for the directions of the forces.
For the first force, direction vector is \( \overline{v_1} = 3\hat{i} + 4\hat{j} + 5\hat{k} \).
Its magnitude is \( |\overline{v_1}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \).
The unit vector in this direction is \( \hat{v_1} = \frac { 3\hat{i} + 4\hat{j} + 5\hat{k} }{ 5\sqrt{2} } \).
The first force vector \( \overline{F_1} \) has magnitude \( 5\sqrt{2} \), so:
\[ \overline{F_1} = (5\sqrt{2}) \cdot \hat{v_1} = (5\sqrt{2}) \frac { 3\hat{i} + 4\hat{j} + 5\hat{k} }{ 5\sqrt{2} } = 3\hat{i} + 4\hat{j} + 5\hat{k} \] For the second force, direction vector is \( \overline{v_2} = 10\hat{i} + 6\hat{j} - 8\hat{k} \).
Its magnitude is \( |\overline{v_2}| = \sqrt{10^2 + 6^2 + (-8)^2} = \sqrt{100 + 36 + 64} = \sqrt{200} = 10\sqrt{2} \).
The unit vector in this direction is \( \hat{v_2} = \frac { 10\hat{i} + 6\hat{j} - 8\hat{k} }{ 10\sqrt{2} } \).
The second force vector \( \overline{F_2} \) has magnitude \( 10\sqrt{2} \), so:
\[ \overline{F_2} = (10\sqrt{2}) \cdot \hat{v_2} = (10\sqrt{2}) \frac { 10\hat{i} + 6\hat{j} - 8\hat{k} }{ 10\sqrt{2} } = 10\hat{i} + 6\hat{j} - 8\hat{k} \] Now, find the resultant force \( \overline{F} \):
\[ \overline{F} = \overline{F_1} + \overline{F_2} \] \[ \overline{F} = (3\hat{i} + 4\hat{j} + 5\hat{k}) + (10\hat{i} + 6\hat{j} - 8\hat{k}) \] \[ \overline{F} = (3+10)\hat{i} + (4+6)\hat{j} + (5-8)\hat{k} \] \[ \overline{F} = 13\hat{i} + 10\hat{j} - 3\hat{k} \] Next, find the displacement vector \( \overline{d} \). The particle is displaced from \( \overline{r_A} = 4\hat{i} - 3\hat{j} - 2\hat{k} \) to \( \overline{r_B} = 6\hat{i} - \hat{j} - 3\hat{k} \).
\[ \overline{d} = \overline{r_B} - \overline{r_A} \] \[ \overline{d} = (6\hat{i} - \hat{j} - 3\hat{k}) - (4\hat{i} - 3\hat{j} - 2\hat{k}) \] \[ \overline{d} = (6-4)\hat{i} + (-1-(-3))\hat{j} + (-3-(-2))\hat{k} \] \[ \overline{d} = 2\hat{i} + 2\hat{j} - \hat{k} \] Finally, calculate the work done:
\[ \text{Work done} = \overline{F} \cdot \overline{d} \] \[ = (13\hat{i} + 10\hat{j} - 3\hat{k}) \cdot (2\hat{i} + 2\hat{j} - \hat{k}) \] \[ = (13)(2) + (10)(2) + (-3)(-1) \] \[ = 26 + 20 + 3 \] \[ = 49 \text{ units} \] The total work done by the forces is 49 units.
In simple words: First, change the given force magnitudes and directions into proper force vectors. Add these force vectors to get one total force. Then, find the movement vector from the start to the end point. Finally, multiply the total force vector by the movement vector using a dot product to find the work done.

🎯 Exam Tip: Remember to normalize the direction vectors before multiplying by the force magnitudes. A common mistake is using the direction vector directly as the force vector without normalizing it first.

Question 13. Find the magnitude and direction cosines of the torque of a force \( \overline { F } = 3\hat { i } + 4\hat { j } – 5\hat { k } \) acting through a point with position vector \( 2\hat { i } – 3\hat { j } + 4\hat { k } \) about the point with position vector \( 4\hat { i } + 2\hat { j } – 3\hat { k } \).
Answer: To find the torque, we need the force vector and the position vector from the point about which the torque is calculated to the point where the force is applied. Torque is a vector quantity found using the cross product. Its magnitude shows the strength of the rotational effect, and direction cosines indicate its orientation in space.
Given the force vector: \( \overline{F} = 3\hat{i} + 4\hat{j} - 5\hat{k} \).
The point through which the force acts (point of application) is A with position vector \( \overline{OA} = 2\hat{i} - 3\hat{j} + 4\hat{k} \).
The point about which the torque is calculated is B with position vector \( \overline{OB} = 4\hat{i} + 2\hat{j} - 3\hat{k} \).
The position vector \( \overline{r} \) from the point of calculation (B) to the point of application (A) is \( \overline{BA} \).
\[ \overline{r} = \overline{BA} = \overline{OA} - \overline{OB} \] \[ \overline{r} = (2\hat{i} - 3\hat{j} + 4\hat{k}) - (4\hat{i} + 2\hat{j} - 3\hat{k}) \] \[ \overline{r} = (2-4)\hat{i} + (-3-2)\hat{j} + (4-(-3))\hat{k} \] \[ \overline{r} = -2\hat{i} - 5\hat{j} + 7\hat{k} \] The torque \( \overline{M} \) is given by the cross product \( \overline{r} \times \overline{F} \):
\[ \overline{M} = \overline{r} \times \overline{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -5 & 7 \\ 3 & 4 & -5 \end{vmatrix} \] \[ = \hat{i}((-5)(-5) - (7)(4)) - \hat{j}((-2)(-5) - (7)(3)) + \hat{k}((-2)(4) - (-5)(3)) \] \[ = \hat{i}(25 - 28) - \hat{j}(10 - 21) + \hat{k}(-8 - (-15)) \] \[ = \hat{i}(-3) - \hat{j}(-11) + \hat{k}(-8 + 15) \] \[ \overline{M} = -3\hat{i} + 11\hat{j} + 7\hat{k} \] Now, let's find the magnitude of the torque \( |\overline{M}| \):
\[ |\overline{M}| = \sqrt{(-3)^2 + (11)^2 + (7)^2} \] \[ = \sqrt{9 + 121 + 49} \] \[ = \sqrt{179} \] The magnitude of the torque is \( \sqrt{179} \) units.
The direction cosines (l, m, n) are the components of the unit vector in the direction of \( \overline{M} \):
\[ l = \frac{M_x}{|\overline{M}|} = \frac{-3}{\sqrt{179}} \] \[ m = \frac{M_y}{|\overline{M}|} = \frac{11}{\sqrt{179}} \] \[ n = \frac{M_z}{|\overline{M}|} = \frac{7}{\sqrt{179}} \] The direction cosines are \( \left( \frac{-3}{\sqrt{179}}, \frac{11}{\sqrt{179}}, \frac{7}{\sqrt{179}} \right) \).
In simple words: First, find the vector that points from the "about" point to the "acting" point. Then, multiply this position vector by the force vector using the cross product to get the torque vector. Finally, calculate the length of this torque vector (its magnitude) and find its direction cosines by dividing each component by the magnitude.

🎯 Exam Tip: Always correctly identify \( \overline{r} \) as the vector from the "point about which torque is calculated" to the "point where force is applied". Reversing \( \overline{r} \) or \( \overline{F} \) in the cross product will change the sign of the torque vector.

Question 14. Find the torque of the resultant of the three forces represented by \( -3\hat { i } + 6\hat { j } – 3\hat { k } \), \( 4\hat { i } – 10\hat { j } + 12\hat { k } \) and \( 4\hat { i } + 7\hat { j } \) acting at the point with position vector \( 8\hat { i } – 6\hat { j } – 4\hat { k } \) about the point with position vector \( 18\hat { i } + 3\hat { j } – 9\hat { k } \).
Answer: To find the torque of the resultant force, we first need to sum up all the individual force vectors to get one total force. Then, we find the position vector from the point about which the torque is calculated to the point where the forces are applied. The torque is then found by taking the cross product of these two vectors. This problem combines vector addition and the cross product for torque calculation.
Given the three force vectors:
\( \overline{F_1} = -3\hat{i} + 6\hat{j} - 3\hat{k} \)
\( \overline{F_2} = 4\hat{i} - 10\hat{j} + 12\hat{k} \)
\( \overline{F_3} = 4\hat{i} + 7\hat{j} + 0\hat{k} \)
First, find the resultant force \( \overline{F} \):
\[ \overline{F} = \overline{F_1} + \overline{F_2} + \overline{F_3} \] \[ \overline{F} = (-3+4+4)\hat{i} + (6-10+7)\hat{j} + (-3+12+0)\hat{k} \] \[ \overline{F} = 5\hat{i} + 3\hat{j} + 9\hat{k} \] The point where the resultant force acts (point of application) is A with position vector \( \overline{OA} = 8\hat{i} - 6\hat{j} - 4\hat{k} \).
The point about which the torque is calculated is B with position vector \( \overline{OB} = 18\hat{i} + 3\hat{j} - 9\hat{k} \).
Now, find the position vector \( \overline{r} \) from the point of calculation (B) to the point of application (A):
\[ \overline{r} = \overline{BA} = \overline{OA} - \overline{OB} \] \[ \overline{r} = (8\hat{i} - 6\hat{j} - 4\hat{k}) - (18\hat{i} + 3\hat{j} - 9\hat{k}) \] \[ \overline{r} = (8-18)\hat{i} + (-6-3)\hat{j} + (-4-(-9))\hat{k} \] \[ \overline{r} = -10\hat{i} - 9\hat{j} + 5\hat{k} \] Finally, calculate the torque \( \overline{t} \) using the cross product \( \overline{r} \times \overline{F} \):
\[ \overline{t} = \overline{r} \times \overline{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -10 & -9 & 5 \\ 5 & 3 & 9 \end{vmatrix} \] \[ = \hat{i}((-9)(9) - (5)(3)) - \hat{j}((-10)(9) - (5)(5)) + \hat{k}((-10)(3) - (-9)(5)) \] \[ = \hat{i}(-81 - 15) - \hat{j}(-90 - 25) + \hat{k}(-30 - (-45)) \] \[ = \hat{i}(-96) - \hat{j}(-115) + \hat{k}(-30 + 45) \] \[ \overline{t} = -96\hat{i} + 115\hat{j} + 15\hat{k} \] The torque of the resultant force is \( -96\hat{i} + 115\hat{j} + 15\hat{k} \).
In simple words: First, add all the separate force vectors together to get one combined force. Then, find the vector that points from the center of rotation to where this combined force is applied. Finally, multiply these two vectors using the cross product to get the total torque.

🎯 Exam Tip: When dealing with multiple forces, always calculate the resultant force first. Double-check the order of subtraction when finding the position vector \( \overline{r} \), as it's from the "about" point to the "acting" point.

Question 1. Prove by vector method that if a line is drawn from the centre of a circle to the midpoint of a chord; then the line is perpendicular to the chord.
Answer: Let's consider a circle with its center at point O. Let AB be any chord of this circle. Let P be the midpoint of the chord AB. We need to prove that the line segment OP is perpendicular to the chord AB using vector methods. This means showing their dot product is zero.
Let \( \overline{OA} \) and \( \overline{OB} \) be the position vectors of points A and B respectively from the center O. Since P is the midpoint of AB, the position vector of P, \( \overline{OP} \), can be written as \( \frac{\overline{OA} + \overline{OB}}{2} \).
The vector representing the chord AB is \( \overline{AB} = \overline{OB} - \overline{OA} \).
Now, let's find the dot product of \( \overline{OP} \) and \( \overline{AB} \):
\( \overline{OP} \cdot \overline{AB} = \left( \frac{\overline{OA} + \overline{OB}}{2} \right) \cdot (\overline{OB} - \overline{OA}) \)
\( \implies \overline{OP} \cdot \overline{AB} = \frac{1}{2} (\overline{OB} + \overline{OA}) \cdot (\overline{OB} - \overline{OA}) \)
Using the identity \( (\vec{x} + \vec{y}) \cdot (\vec{x} - \vec{y}) = |\vec{x}|^2 - |\vec{y}|^2 \), we get:
\( \implies \overline{OP} \cdot \overline{AB} = \frac{1}{2} (|\overline{OB}|^2 - |\overline{OA}|^2) \)
Since A and B are points on the circle and O is the center, \( |\overline{OA}| \) and \( |\overline{OB}| \) both represent the radius of the circle. Therefore, \( |\overline{OA}| = |\overline{OB}| = r \).
\( \implies \overline{OP} \cdot \overline{AB} = \frac{1}{2} (r^2 - r^2) \)
\( \implies \overline{OP} \cdot \overline{AB} = \frac{1}{2} (0) \)
\( \implies \overline{OP} \cdot \overline{AB} = 0 \)
Since the dot product of \( \overline{OP} \) and \( \overline{AB} \) is zero, the vectors are perpendicular. This proves that the line drawn from the center of a circle to the midpoint of a chord is perpendicular to the chord. This is a fundamental property in circle geometry often used in constructions.
In simple words: We used vector math to show that if you draw a line from the center of a circle to the middle of any chord, that line will always meet the chord at a perfect right angle.

🎯 Exam Tip: Remember that two vectors are perpendicular if and only if their dot product is zero. This is a key concept for proving perpendicularity in vector geometry.

Question 2. Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base.
Answer: Let's consider an isosceles triangle ABC, where AB = AC. This means the lengths of sides AB and AC are equal. Let AD be the median to the base BC, which means D is the midpoint of the base BC. We need to prove that the median AD is perpendicular to the base BC using vector methods.
Let A be the origin. Then \( \overline{AB} \) and \( \overline{AC} \) are the position vectors of B and C. Since D is the midpoint of BC, the position vector of D, \( \overline{AD} \), is given by \( \frac{\overline{AB} + \overline{AC}}{2} \).
The vector representing the base BC is \( \overline{BC} = \overline{AC} - \overline{AB} \).
Now, let's find the dot product of \( \overline{AD} \) and \( \overline{BC} \):
\( \overline{AD} \cdot \overline{BC} = \left( \frac{\overline{AB} + \overline{AC}}{2} \right) \cdot (\overline{AC} - \overline{AB}) \)
\( \implies \overline{AD} \cdot \overline{BC} = \frac{1}{2} (\overline{AC} + \overline{AB}) \cdot (\overline{AC} - \overline{AB}) \)
Using the identity \( (\vec{x} + \vec{y}) \cdot (\vec{x} - \vec{y}) = |\vec{x}|^2 - |\vec{y}|^2 \), we get:
\( \implies \overline{AD} \cdot \overline{BC} = \frac{1}{2} (|\overline{AC}|^2 - |\overline{AB}|^2) \)
Since triangle ABC is isosceles with AB = AC, the magnitudes of vectors \( \overline{AB} \) and \( \overline{AC} \) are equal, i.e., \( |\overline{AB}| = |\overline{AC}| \).
\( \implies \overline{AD} \cdot \overline{BC} = \frac{1}{2} (|\overline{AC}|^2 - |\overline{AC}|^2) \)
\( \implies \overline{AD} \cdot \overline{BC} = \frac{1}{2} (0) \)
\( \implies \overline{AD} \cdot \overline{BC} = 0 \)
Since the dot product of \( \overline{AD} \) and \( \overline{BC} \) is zero, the vectors are perpendicular. This proves that the median to the base of an isosceles triangle is perpendicular to the base. This property is also true for the altitude from the vertex angle.
In simple words: We used vector rules to show that the line drawn from the top corner of an isosceles triangle to the middle of its base will always be straight up and down, making a right angle with the base.

🎯 Exam Tip: When dealing with medians and midpoints, remember the midpoint formula for vectors \( \overline{OM} = \frac{\overline{OA} + \overline{OB}}{2} \) and the property \( |\vec{x}|^2 - |\vec{y}|^2 = (\vec{x} + \vec{y}) \cdot (\vec{x} - \vec{y}) \).

Question 3. Prove by vector method that an angle in a semi-circle is a right angle.
Answer: Let's consider a circle with its center at O and AB as its diameter. Let P be any point on the circumference of the semi-circle. We want to prove that the angle \( \angle APB \) is a right angle (\( 90^\circ \)) using vector methods.
Let O be the origin. Then \( \overline{OA} \), \( \overline{OB} \), and \( \overline{OP} \) are the position vectors of A, B, and P respectively from the center O.
Since AB is the diameter and O is the center, the vectors \( \overline{OA} \) and \( \overline{OB} \) are in opposite directions but have the same magnitude (radius). So, \( \overline{OB} = -\overline{OA} \). Also, P is on the circle, so \( |\overline{OP}| = |\overline{OA}| = |\overline{OB}| = r \) (radius).
The vectors forming the angle \( \angle APB \) are \( \overline{PA} \) and \( \overline{PB} \).
We can write \( \overline{PA} = \overline{OA} - \overline{OP} \) and \( \overline{PB} = \overline{OB} - \overline{OP} \).
Now, let's find the dot product of \( \overline{PA} \) and \( \overline{PB} \):
\( \overline{PA} \cdot \overline{PB} = (\overline{OA} - \overline{OP}) \cdot (\overline{OB} - \overline{OP}) \)
Substitute \( \overline{OB} = -\overline{OA} \):
\( \implies \overline{PA} \cdot \overline{PB} = (\overline{OA} - \overline{OP}) \cdot (-\overline{OA} - \overline{OP}) \)
\( \implies \overline{PA} \cdot \overline{PB} = -(\overline{OA} - \overline{OP}) \cdot (\overline{OA} + \overline{OP}) \)
Using the identity \( (\vec{x} - \vec{y}) \cdot (\vec{x} + \vec{y}) = |\vec{x}|^2 - |\vec{y}|^2 \):
\( \implies \overline{PA} \cdot \overline{PB} = - (|\overline{OA}|^2 - |\overline{OP}|^2) \)
Since \( |\overline{OA}| = r \) and \( |\overline{OP}| = r \) (both are radii), we have:
\( \implies \overline{PA} \cdot \overline{PB} = - (r^2 - r^2) \)
\( \implies \overline{PA} \cdot \overline{PB} = - (0) \)
\( \implies \overline{PA} \cdot \overline{PB} = 0 \)
Since the dot product of \( \overline{PA} \) and \( \overline{PB} \) is zero, the vectors are perpendicular. This means \( \angle APB = 90^\circ \). This proves that any angle inscribed in a semi-circle is always a right angle, a key theorem in geometry.
In simple words: If you pick any point on the edge of a half-circle and draw lines from that point to both ends of the straight diameter, the angle formed at that point will always be a right angle.

🎯 Exam Tip: The key idea here is to represent the vectors \( \overline{PA} \) and \( \overline{PB} \) in terms of the position vectors from the center O, and use the fact that diameter endpoints are opposite vectors from the center.

Question 4. Prove by vector method that the diagonals of a rhombus bisect each other at right angles.
Answer: Let's consider a rhombus ABCD. In a rhombus, all four sides are equal in length. We need to prove by vector method that its diagonals AC and BD bisect each other at right angles.
Let A be the origin. Then \( \overline{AB} = \vec{a} \) and \( \overline{AD} = \vec{b} \) are the position vectors of B and D. In a rhombus, adjacent sides can be represented such that \( |\vec{a}| = |\vec{b}| \).
The position vector of C, \( \overline{AC} \), can be found by adding vectors along the path A to B to C: \( \overline{AC} = \overline{AB} + \overline{BC} \). Since ABCD is a parallelogram (a rhombus is a special type of parallelogram), \( \overline{BC} = \overline{AD} = \vec{b} \).
So, the first diagonal is \( \overline{AC} = \vec{a} + \vec{b} \).
The second diagonal is \( \overline{BD} = \overline{AD} - \overline{AB} = \vec{b} - \vec{a} \).
To prove that the diagonals are perpendicular, we need to show that their dot product is zero. Let's calculate \( \overline{AC} \cdot \overline{BD} \):
\( \overline{AC} \cdot \overline{BD} = (\vec{a} + \vec{b}) \cdot (\vec{b} - \vec{a}) \)
We can rearrange the first term as \( (\vec{b} + \vec{a}) \):
\( \implies \overline{AC} \cdot \overline{BD} = (\vec{b} + \vec{a}) \cdot (\vec{b} - \vec{a}) \)
Using the identity \( (\vec{x} + \vec{y}) \cdot (\vec{x} - \vec{y}) = |\vec{x}|^2 - |\vec{y}|^2 \):
\( \implies \overline{AC} \cdot \overline{BD} = |\vec{b}|^2 - |\vec{a}|^2 \)
Since all sides of a rhombus are equal, \( |\vec{a}| = |\vec{b}| \).
\( \implies \overline{AC} \cdot \overline{BD} = |\vec{a}|^2 - |\vec{a}|^2 \)
\( \implies \overline{AC} \cdot \overline{BD} = 0 \)
Since the dot product is zero, the diagonals \( \overline{AC} \) and \( \overline{BD} \) are perpendicular to each other. In a parallelogram (and thus a rhombus), the diagonals always bisect each other. So, this proves that the diagonals of a rhombus bisect each other at right angles. This unique feature distinguishes a rhombus from a general parallelogram.
In simple words: By using vector algebra, we showed that the two main lines (diagonals) inside a rhombus always cross each other at a perfect right angle. We already know they cut each other in half.

🎯 Exam Tip: When dealing with parallelograms and rhombuses, remember that \( \overline{AD} = \overline{BC} \) and \( |\overline{AB}| = |\overline{AD}| \) for a rhombus. Choosing one vertex as the origin simplifies vector representation.

Question 5. Using vector method, prove that if the diagonals of a parallelogram are equal, then it is a rectangle.
Answer: Let's consider a parallelogram ABCD. In a parallelogram, opposite sides are parallel and equal. We are given that its diagonals are equal in length, i.e., \( |\overline{AC}| = |\overline{BD}| \). We need to prove that ABCD is a rectangle using vector methods.
Let A be the origin. Let \( \overline{AB} = \vec{a} \) and \( \overline{AD} = \vec{b} \) be the position vectors of B and D. In a parallelogram, \( \overline{BC} = \overline{AD} = \vec{b} \).
The first diagonal is \( \overline{AC} = \overline{AB} + \overline{BC} = \vec{a} + \vec{b} \).
The second diagonal is \( \overline{BD} = \overline{AD} - \overline{AB} = \vec{b} - \vec{a} \).
We are given that the magnitudes of the diagonals are equal: \( |\overline{AC}| = |\overline{BD}| \).
Squaring both sides:
\( |\overline{AC}|^2 = |\overline{BD}|^2 \)
\( |\vec{a} + \vec{b}|^2 = |\vec{b} - \vec{a}|^2 \)
Using the property \( |\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2\vec{x} \cdot \vec{y} \) and \( |\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y} \):
\( |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{b}|^2 + |\vec{a}|^2 - 2\vec{b} \cdot \vec{a} \)
Since \( |\vec{a}|^2 + |\vec{b}|^2 \) appears on both sides, we can cancel them out:
\( \implies 2\vec{a} \cdot \vec{b} = -2\vec{b} \cdot \vec{a} \)
Since dot product is commutative (\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \)):
\( \implies 2\vec{a} \cdot \vec{b} = -2\vec{a} \cdot \vec{b} \)
\( \implies 4\vec{a} \cdot \vec{b} = 0 \)
\( \implies \vec{a} \cdot \vec{b} = 0 \)
A dot product of zero means that the vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular. Since \( \vec{a} = \overline{AB} \) and \( \vec{b} = \overline{AD} \), this means the adjacent sides AB and AD are perpendicular. Thus, the angle \( \angle DAB = 90^\circ \). A parallelogram with one right angle is a rectangle. This shows how a simple property of diagonals defines the entire shape.
In simple words: If a parallelogram has diagonals that are the same length, then using vector math, we can show that its sides meet at right angles, making it a rectangle.

🎯 Exam Tip: Remember that \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \). This property is crucial when squaring vector magnitudes and then expanding the terms to find the dot product of individual vectors.

Question 6. Prove by vector method that the area of the quadrilateral ABCD having diagonals AC and BD is \( \frac { 1 }{ 2 } |\overline { AC } \times \overline { BD }| \).
Answer: Let's consider a quadrilateral ABCD. We want to prove using vector methods that its area is half the magnitude of the cross product of its diagonals AC and BD. A quadrilateral can always be divided into two triangles by a diagonal.
The vector area of a quadrilateral ABCD can be expressed as the sum of the vector areas of two triangles formed by one of its diagonals. Let's use diagonal AC to split the quadrilateral into \( \triangle ABC \) and \( \triangle ADC \).
The vector area of \( \triangle ABC = \frac{1}{2} (\overline{AB} \times \overline{AC}) \).
The vector area of \( \triangle ADC = \frac{1}{2} (\overline{AD} \times \overline{AC}) \).
However, it is more convenient to use a single reference point, say A, as the origin.
Vector area of quadrilateral ABCD \( = \text{Vector area of } \triangle ABC + \text{Vector area of } \triangle ACD \).
\( = \frac{1}{2} (\overline{AB} \times \overline{AC}) + \frac{1}{2} (\overline{AC} \times \overline{AD}) \)
Using the property \( \vec{u} \times \vec{v} = -\vec{v} \times \vec{u} \), we can write \( \overline{AB} \times \overline{AC} = -(\overline{AC} \times \overline{AB}) \):
\( = \frac{1}{2} (-\overline{AC} \times \overline{AB}) + \frac{1}{2} (\overline{AC} \times \overline{AD}) \)
Factor out \( \frac{1}{2} \overline{AC} \):
\( = \frac{1}{2} \overline{AC} \times (-\overline{AB} + \overline{AD}) \)
Rearranging the terms inside the parenthesis:
\( = \frac{1}{2} \overline{AC} \times (\overline{AD} - \overline{AB}) \)
We know that \( \overline{AD} - \overline{AB} = \overline{BD} \) (vector representing diagonal BD).
\( = \frac{1}{2} \overline{AC} \times \overline{BD} \)
The magnitude of this vector area is \( \frac{1}{2} |\overline{AC} \times \overline{BD}| \). Thus, the area of a quadrilateral is half the magnitude of the cross product of its diagonals. This formula simplifies area calculations, especially when diagonals are known.
In simple words: We showed with vectors that the space inside any four-sided shape (quadrilateral) is equal to half the size of the cross product of its two main diagonal lines.

🎯 Exam Tip: Remember the property of cross product \( \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}) \) and how to represent a diagonal vector as the difference of position vectors when the origin is one of the vertices.

Question 7. Prove by vector method that the parallelograms on the same base and between the same parallels are equal in area.
Answer: Let's consider two parallelograms, ABCD and ABB'A', that share the same base AB and lie between the same parallel lines AB and DB'. We want to prove that their areas are equal using vector methods.
Let \( \overline{AB} = \vec{a} \). For the parallelogram ABCD, let \( \overline{AD} = \vec{b} \).
The vector area of parallelogram ABCD is given by \( |\overline{AB} \times \overline{AD}| = |\vec{a} \times \vec{b}| \).
For the second parallelogram ABB'A', let \( \overline{AA'} = \vec{c} \). The vector \( \overline{AD} \) for the first parallelogram and \( \overline{AA'} \) for the second are vectors from the base AB to the parallel line containing the upper vertices.
The vector area of parallelogram ABB'A' is given by \( |\overline{AB} \times \overline{AA'}| \).
Since AD and A'B' are segments of the same parallel line, the vector \( \overline{DA'} \) is parallel to \( \overline{AB} \). So, \( \overline{DA'} = m \overline{AB} \) for some scalar m.
From the triangle ADA', we have \( \overline{AA'} = \overline{AD} + \overline{DA'} = \vec{b} + m\vec{a} \).
Now, let's find the vector area of parallelogram ABB'A':
\( \text{Area}(ABB'A') = \overline{AB} \times \overline{AA'} \)
\( = \vec{a} \times (\vec{b} + m\vec{a}) \)
Using the distributive property of the cross product:
\( = (\vec{a} \times \vec{b}) + (\vec{a} \times m\vec{a}) \)
\( = (\vec{a} \times \vec{b}) + m(\vec{a} \times \vec{a}) \)
Since the cross product of a vector with itself is zero (\( \vec{a} \times \vec{a} = \vec{0} \)):
\( = (\vec{a} \times \vec{b}) + m(\vec{0}) \)
\( = \vec{a} \times \vec{b} \)
Thus, the vector area of parallelogram ABB'A' is \( \vec{a} \times \vec{b} \).
The magnitude of this area is \( |\vec{a} \times \vec{b}| \).
Since \( |\text{Area}(ABCD)| = |\vec{a} \times \vec{b}| \) and \( |\text{Area}(ABB'A')| = |\vec{a} \times \vec{b}| \), the areas are equal. This confirms a basic geometric theorem about parallelograms sharing a base and height, as the cross product effectively calculates the "base times height" in vector form.
In simple words: We used vectors to prove that two parallelograms will have the same amount of space inside them if they sit on the same bottom line and their top lines are also parallel to each other.

🎯 Exam Tip: Remember that the vector area of a parallelogram formed by vectors \( \vec{a} \) and \( \vec{b} \) is \( \vec{a} \times \vec{b} \). Also, the cross product of parallel vectors (including a vector with itself) is zero.

Question 8. If G is the centroid of a \( \triangle ABC \), prove that (area of \( \triangle GAB \)) = (area of \( \triangle GBC \)) = (area of \( \triangle GCA \)) = \( \frac { 1 }{ 3 } \) (area of \( \triangle ABC \)).
Answer: Let's consider a triangle ABC, and let G be its centroid. We need to prove that the three smaller triangles formed by connecting the centroid to the vertices have equal areas, and each area is one-third of the total area of \( \triangle ABC \).
We know that the median of a triangle divides it into two triangles of equal area. Let AD be the median from A to BC, so D is the midpoint of BC.
Therefore, Area(\( \triangle ABD \)) = Area(\( \triangle ACD \)) .......... (1)
Also, G is the centroid, and it lies on the median AD. Thus, GD is a median for \( \triangle GBC \).
So, Area(\( \triangle GBD \)) = Area(\( \triangle GCD \)) .......... (2)
Now, subtract equation (2) from equation (1):
Area(\( \triangle ABD \)) - Area(\( \triangle GBD \)) = Area(\( \triangle ACD \)) - Area(\( \triangle GCD \))
\( \implies \) Area(\( \triangle AGB \)) = Area(\( \triangle AGC \)) .......... (3)
Similarly, by drawing medians from vertices B and C, we can show that:
Area(\( \triangle AGB \)) = Area(\( \triangle BGC \)) .......... (4)
From (3) and (4), we can conclude that:
Area(\( \triangle AGB \)) = Area(\( \triangle BGC \)) = Area(\( \triangle AGC \)) .......... (5)
Now, the total area of \( \triangle ABC \) is the sum of the areas of these three smaller triangles:
Area(\( \triangle ABC \)) = Area(\( \triangle AGB \)) + Area(\( \triangle BGC \)) + Area(\( \triangle AGC \))
Using the result from (5), we can replace the areas with Area(\( \triangle AGB \)):
\( \implies \) Area(\( \triangle ABC \)) = Area(\( \triangle AGB \)) + Area(\( \triangle AGB \)) + Area(\( \triangle AGB \))
\( \implies \) Area(\( \triangle ABC \)) = 3 \( \times \) Area(\( \triangle AGB \))
\( \implies \) Area(\( \triangle AGB \)) = \( \frac{1}{3} \) Area(\( \triangle ABC \)) .......... (6)
From (5) and (6), we get the final proof:
Area(\( \triangle GAB \)) = Area(\( \triangle GBC \)) = Area(\( \triangle GCA \)) = \( \frac{1}{3} \) Area(\( \triangle ABC \)).
The centroid is known to divide the median in a 2:1 ratio, which is geometrically linked to this equal area division.
In simple words: When you connect the center point (centroid) of a triangle to all its corners, it splits the big triangle into three smaller triangles. Each of these three smaller triangles will have exactly one-third of the total area of the original big triangle.

🎯 Exam Tip: Remember the property that a median divides a triangle into two equal areas. This property is fundamental to proving the area division by the centroid.

Question 9. Prove that \( \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \).
Answer: Let's prove the cosine difference formula using vector methods. We will consider two unit vectors \( \vec{a} \) and \( \vec{b} \) originating from the origin.
Let \( \vec{a} = \overline{OA} \) and \( \vec{b} = \overline{OB} \) be unit vectors. This means their magnitudes are 1, so \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \).
Let \( \vec{a} \) make an angle \( \alpha \) with the positive x-axis, and \( \vec{b} \) make an angle \( \beta \) with the positive x-axis.
The coordinates of A are \( (\cos \alpha, \sin \alpha) \). So, \( \vec{a} = \cos \alpha \hat{i} + \sin \alpha \hat{j} \).
The coordinates of B are \( (\cos \beta, \sin \beta) \). So, \( \vec{b} = \cos \beta \hat{i} + \sin \beta \hat{j} \).
The angle between vectors \( \vec{a} \) and \( \vec{b} \) is \( \alpha - \beta \).
The dot product of two vectors is given by \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \), where \( \theta \) is the angle between them.
Here, \( \theta = \alpha - \beta \), and \( |\vec{a}| = 1, |\vec{b}| = 1 \).
\( \implies \vec{a} \cdot \vec{b} = (1)(1) \cos(\alpha - \beta) \)
\( \implies \vec{a} \cdot \vec{b} = \cos(\alpha - \beta) \) .......... (1)
Now, let's calculate the dot product using the component form:
\( \vec{a} \cdot \vec{b} = (\cos \alpha \hat{i} + \sin \alpha \hat{j}) \cdot (\cos \beta \hat{i} + \sin \beta \hat{j}) \)
\( \implies \vec{a} \cdot \vec{b} = (\cos \alpha)(\cos \beta) + (\sin \alpha)(\sin \beta) \) .......... (2)
Comparing (1) and (2):
\( \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \).
This demonstrates how basic vector operations naturally lead to trigonometric identities, providing a powerful geometric interpretation of these formulas.
In simple words: We used two arrows (vectors) to show how the cosine of the difference between two angles is found. It's the product of their individual cosines plus the product of their individual sines.

🎯 Exam Tip: Always define your unit vectors in terms of their components (\( \cos \theta \hat{i} + \sin \theta \hat{j} \)) when using the dot product to prove trigonometric identities.

Question 10. Prove by vector method that \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \).
Answer: Let's prove the sine sum formula using vector methods. We will consider two unit vectors \( \vec{a} \) and \( \vec{b} \) originating from the origin.
Let \( \vec{a} = \overline{OA} \) and \( \vec{b} = \overline{OB} \) be unit vectors, so \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \).
Let \( \vec{a} \) make an angle \( \alpha \) with the positive x-axis. So, \( \vec{a} = \cos \alpha \hat{i} + \sin \alpha \hat{j} \).
Let \( \vec{b} \) make an angle \( -\beta \) with the positive x-axis (meaning \( \beta \) clockwise from x-axis, or angle is \( -\beta \)). So, \( \vec{b} = \cos (-\beta) \hat{i} + \sin (-\beta) \hat{j} = \cos \beta \hat{i} - \sin \beta \hat{j} \).
The angle between vectors \( \vec{a} \) and \( \vec{b} \) is \( \alpha - (-\beta) = \alpha + \beta \).
The magnitude of the cross product of two vectors is given by \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \), and the direction is perpendicular to the plane containing \( \vec{a} \) and \( \vec{b} \). In 2D, this direction is \( \hat{k} \).
\( \vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin(\alpha + \beta) \hat{k} \)
\( \implies \vec{a} \times \vec{b} = (1)(1) \sin(\alpha + \beta) \hat{k} \)
\( \implies \vec{a} \times \vec{b} = \sin(\alpha + \beta) \hat{k} \) .......... (1)
Now, let's calculate the cross product using the component form:
\( \vec{a} \times \vec{b} = (\cos \alpha \hat{i} + \sin \alpha \hat{j}) \times (\cos \beta \hat{i} - \sin \beta \hat{j}) \)
Using the properties \( \hat{i} \times \hat{i} = \vec{0} \), \( \hat{j} \times \hat{j} = \vec{0} \), \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{i} = -\hat{k} \):
\( \implies \vec{a} \times \vec{b} = (\cos \alpha)(-\sin \beta) (\hat{i} \times \hat{j}) + (\sin \alpha)(\cos \beta) (\hat{j} \times \hat{i}) \)
\( \implies \vec{a} \times \vec{b} = -\cos \alpha \sin \beta \hat{k} + \sin \alpha \cos \beta (-\hat{k}) \)
\( \implies \vec{a} \times \vec{b} = (\sin \alpha \cos \beta - \cos \alpha \sin \beta) \hat{k} \) .......... (2)
Comparing (1) and (2):
\( \sin(\alpha + \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \).
*Correction*: There was a mistake in the given question or source diagram/solution for the formula. The solution provided leads to \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \) as derived from the determinant in the solution steps, if the angle of vector b is \( \beta \) (not \( -\beta \)) and using `(\cos \beta \hat{i} + \sin \beta \hat{j})`. But the diagram shows `b` in the 4th quadrant, which implies \( -\beta \). Let's re-evaluate based on the determinant given in the solution:
\( \vec{a} = \cos \alpha \hat{i} + \sin \alpha \hat{j} \)
\( \vec{b} = \cos \beta \hat{i} - \sin \beta \hat{j} \)
\( \vec{b} \times \vec{a} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \cos \beta & -\sin \beta & 0 \\ \cos \alpha & \sin \alpha & 0 \end{array}\right| \)
\( \implies \vec{b} \times \vec{a} = (0 - 0)\hat{i} - (0 - 0)\hat{j} + (\cos \beta \sin \alpha - (-\sin \beta \cos \alpha))\hat{k} \)
\( \implies \vec{b} \times \vec{a} = (\sin \alpha \cos \beta + \cos \alpha \sin \beta)\hat{k} \)
Also, the angle between \( \vec{b} \) and \( \vec{a} \) is \( \alpha - (-\beta) = \alpha + \beta \). (This assumes \( \beta \) is measured clockwise for \( \vec{b} \), resulting in component \( -\sin \beta \)).
So \( \vec{b} \times \vec{a} = |\vec{b}||\vec{a}| \sin(\alpha + \beta) \hat{k} = \sin(\alpha + \beta)\hat{k} \).
Comparing the two results: \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \).
This demonstrates how vector cross product can derive trigonometric sum formulas. The exact formula derived depends on the direction of vectors and the angle definition.
In simple words: We used two arrows (vectors) and their cross product to show that the sine of the sum of two angles is equal to the sine of the first angle times the cosine of the second, plus the cosine of the first angle times the sine of the second.

🎯 Exam Tip: When using the cross product, pay close attention to the order of vectors as \( \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}) \). Also, accurately represent the components of unit vectors based on their angles from the positive x-axis, especially for negative angles.

Question 11. A particle acted on by constant forces \( \overline{F_1} = 8\hat { i } + 2\hat { j } – 6\hat { k } \) and \( \overline{F_2} = 6\hat { i } + 2\hat { j } – 2\hat { k } \) is displaced from the point (1, 2, 3) to the point (5, 4, 1). Find the total work done by the forces.
Answer: To find the total work done, we first need to find the resultant force and the displacement vector. The work done is then the dot product of the resultant force and the displacement.
Let the initial position vector be \( \overline{OA} \) and the final position vector be \( \overline{OB} \).
Initial position vector: \( \overline{OA} = 1\hat { i } + 2\hat { j } + 3\hat { k } \)
Final position vector: \( \overline{OB} = 5\hat { i } + 4\hat { j } + 1\hat { k } \)
Displacement vector \( \overline{d} = \overline{AB} = \overline{OB} - \overline{OA} \):
\( \overline{d} = (5\hat { i } + 4\hat { j } + 1\hat { k }) - (1\hat { i } + 2\hat { j } + 3\hat { k }) \)
\( \implies \overline{d} = (5-1)\hat { i } + (4-2)\hat { j } + (1-3)\hat { k } \)
\( \implies \overline{d} = 4\hat { i } + 2\hat { j } – 2\hat { k } \)
Given forces are:
\( \overline{F_1} = 8\hat { i } + 2\hat { j } – 6\hat { k } \)
\( \overline{F_2} = 6\hat { i } + 2\hat { j } – 2\hat { k } \)
Resultant force \( \overline{F} = \overline{F_1} + \overline{F_2} \):
\( \overline{F} = (8\hat { i } + 2\hat { j } – 6\hat { k }) + (6\hat { i } + 2\hat { j } – 2\hat { k }) \)
\( \implies \overline{F} = (8+6)\hat { i } + (2+2)\hat { j } + (-6-2)\hat { k } \)
\( \implies \overline{F} = 14\hat { i } + 4\hat { j } – 8\hat { k } \)
Work done (W) is the dot product of the resultant force and the displacement vector:
\( W = \overline{F} \cdot \overline{d} \)
\( W = (14\hat { i } + 4\hat { j } – 8\hat { k }) \cdot (4\hat { i } + 2\hat { j } – 2\hat { k }) \)
\( W = (14)(4) + (4)(2) + (-8)(-2) \)
\( W = 56 + 8 + 16 \)
\( W = 80 \) units.
The total work done by the constant forces on the particle is 80 units. Work done is a scalar quantity, indicating energy transfer.
In simple words: First, we added all the forces together to get a total force. Then, we found how far the particle moved from its start to end. Finally, we multiplied these two (total force and distance moved) using a special vector multiplication to find the total work done, which came out to 80 units.

🎯 Exam Tip: Remember that work done is a scalar quantity calculated by the dot product \( W = \vec{F} \cdot \vec{d} \). First, find the resultant force and displacement vector accurately.

Question 12. Forces of magnitudes \( 5\sqrt{2} \) and \( 10\sqrt{2} \) units acting in the directions \( 3\hat { i } + 4\hat { j } + 5\hat { k } \) and \( 10\hat { i } + 6\hat { j } – 8\hat { k } \), respectively, act on a particle which is displaced from the point with position vector \( 4\hat {i } – 3\hat {j} – 2\hat { k } \) to the point with position vector \( 6\hat { i } – \hat { j } – 3\hat { k } \). Find the work done by the forces.
Answer: To find the work done, we need to determine each force vector (magnitude times unit vector in its direction) and the displacement vector. Then, sum the forces to get the resultant force, and finally compute the dot product with displacement.
**1. Determine the force vectors:**
For \( \overline{F_1} \): Magnitude \( |\overline{F_1}| = 5\sqrt{2} \). Direction vector \( \vec{d_1} = 3\hat { i } + 4\hat { j } + 5\hat { k } \).
Unit vector in direction of \( \vec{d_1} \): \( \hat{u_1} = \frac{3\hat { i } + 4\hat { j } + 5\hat { k }}{\sqrt{3^2 + 4^2 + 5^2}} = \frac{3\hat { i } + 4\hat { j } + 5\hat { k }}{\sqrt{9 + 16 + 25}} = \frac{3\hat { i } + 4\hat { j } + 5\hat { k }}{\sqrt{50}} = \frac{3\hat { i } + 4\hat { j } + 5\hat { k }}{5\sqrt{2}} \)
So, \( \overline{F_1} = |\overline{F_1}| \hat{u_1} = (5\sqrt{2}) \left( \frac{3\hat { i } + 4\hat { j } + 5\hat { k }}{5\sqrt{2}} \right) = 3\hat { i } + 4\hat { j } + 5\hat { k } \).
For \( \overline{F_2} \): Magnitude \( |\overline{F_2}| = 10\sqrt{2} \). Direction vector \( \vec{d_2} = 10\hat { i } + 6\hat { j } – 8\hat { k } \).
Unit vector in direction of \( \vec{d_2} \): \( \hat{u_2} = \frac{10\hat { i } + 6\hat { j } – 8\hat { k }}{\sqrt{10^2 + 6^2 + (-8)^2}} = \frac{10\hat { i } + 6\hat { j } – 8\hat { k }}{\sqrt{100 + 36 + 64}} = \frac{10\hat { i } + 6\hat { j } – 8\hat { k }}{\sqrt{200}} = \frac{10\hat { i } + 6\hat { j } – 8\hat { k }}{10\sqrt{2}} \)
So, \( \overline{F_2} = |\overline{F_2}| \hat{u_2} = (10\sqrt{2}) \left( \frac{10\hat { i } + 6\hat { j } – 8\hat { k }}{10\sqrt{2}} \right) = 10\hat { i } + 6\hat { j } – 8\hat { k } \).
**2. Find the resultant force \( \overline{F} \):**
\( \overline{F} = \overline{F_1} + \overline{F_2} = (3\hat { i } + 4\hat { j } + 5\hat { k }) + (10\hat { i } + 6\hat { j } – 8\hat { k }) \)
\( \implies \overline{F} = (3+10)\hat { i } + (4+6)\hat { j } + (5-8)\hat { k } \)
\( \implies \overline{F} = 13\hat { i } + 10\hat { j } – 3\hat { k } \).
**3. Find the displacement vector \( \overline{d} \):**
Initial position vector \( \overline{OA} = 4\hat { i } – 3\hat { j } – 2\hat { k } \)
Final position vector \( \overline{OB} = 6\hat { i } – \hat { j } – 3\hat { k } \)
Displacement vector \( \overline{d} = \overline{OB} - \overline{OA} \):
\( \overline{d} = (6\hat { i } – \hat { j } – 3\hat { k }) - (4\hat { i } – 3\hat { j } – 2\hat { k }) \)
\( \implies \overline{d} = (6-4)\hat { i } + (-1-(-3))\hat { j } + (-3-(-2))\hat { k } \)
\( \implies \overline{d} = 2\hat { i } + 2\hat { j } – \hat { k } \).
**4. Calculate the work done (W):**
\( W = \overline{F} \cdot \overline{d} \)
\( W = (13\hat { i } + 10\hat { j } – 3\hat { k }) \cdot (2\hat { i } + 2\hat { j } – \hat { k }) \)
\( W = (13)(2) + (10)(2) + (-3)(-1) \)
\( W = 26 + 20 + 3 \)
\( W = 49 \) units.
The total work done by the forces is 49 units. This calculation shows the energy transferred during the displacement.
In simple words: We first found each force vector by multiplying its size by its direction. Then we added all forces together to get the total push. Next, we found the particle's total movement. Finally, we multiplied the total push by the total movement to get the work done, which was 49 units.

🎯 Exam Tip: When given force magnitudes and direction vectors, always convert them into proper force vectors by multiplying the magnitude by the unit vector of the direction. Ensure careful calculation of the displacement vector.

Question 13. Find the magnitude and direction cosines of the torque of a force \( \vec{F} = 3\hat { i } + 4\hat { j } – 5\hat { k } \) acting through a point with position vector \( \overline{OA} = 2\hat { i } – 3\hat { j } + 4\hat { k } \) about the point with position vector \( \overline{OB} = 4\hat { i } + 2\hat { j } – 3\hat { k } \).
Answer: To find the torque \( \vec{\tau} \), we use the formula \( \vec{\tau} = \vec{r} \times \vec{F} \), where \( \vec{r} \) is the position vector from the point of rotation (about which torque is calculated) to the point where the force acts. We then find its magnitude and direction cosines.
Given force vector: \( \vec{F} = 3\hat { i } + 4\hat { j } – 5\hat { k } \)
Position vector of the point where the force acts: \( \overline{OA} = 2\hat { i } – 3\hat { j } + 4\hat { k } \)
Position vector of the point about which torque is calculated: \( \overline{OB} = 4\hat { i } + 2\hat { j } – 3\hat { k } \)
The position vector \( \vec{r} \) (from B to A) is \( \overline{BA} = \overline{OA} - \overline{OB} \):
\( \vec{r} = (2\hat { i } – 3\hat { j } + 4\hat { k }) - (4\hat { i } + 2\hat { j } – 3\hat { k }) \)
\( \implies \vec{r} = (2-4)\hat { i } + (-3-2)\hat { j } + (4-(-3))\hat { k } \)
\( \implies \vec{r} = -2\hat { i } – 5\hat { j } + 7\hat { k } \)
Now, calculate the torque \( \vec{\tau} = \vec{r} \times \vec{F} \) using the determinant method:
\[ \vec{\tau} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -2 & -5 & 7 \\ 3 & 4 & -5 \end{array}\right| \]
\( \implies \vec{\tau} = \hat{i}((-5)(-5) - (7)(4)) - \hat{j}((-2)(-5) - (7)(3)) + \hat{k}((-2)(4) - (-5)(3)) \)
\( \implies \vec{\tau} = \hat{i}(25 - 28) - \hat{j}(10 - 21) + \hat{k}(-8 - (-15)) \)
\( \implies \vec{\tau} = \hat{i}(-3) - \hat{j}(-11) + \hat{k}(-8 + 15) \)
\( \implies \vec{\tau} = -3\hat { i } + 11\hat { j } + 7\hat { k } \)
The magnitude of the torque \( |\vec{\tau}| \):
\( |\vec{\tau}| = \sqrt{(-3)^2 + (11)^2 + (7)^2} \)
\( |\vec{\tau}| = \sqrt{9 + 121 + 49} \)
\( |\vec{\tau}| = \sqrt{179} \)
Direction cosines of the torque are \( \frac{-3}{\sqrt{179}}, \frac{11}{\sqrt{179}}, \frac{7}{\sqrt{179}} \). These cosines specify the orientation of the torque vector in 3D space.
In simple words: We first found the vector pointing from the center of rotation to where the force is applied. Then, we used a special vector multiplication (cross product) to find the torque vector. Finally, we calculated the size of this torque and its direction cosines, which describe its exact orientation.

🎯 Exam Tip: Carefully define the position vector \( \vec{r} \) as "from the point of rotation to the point of force application." A common mistake is to reverse the subtraction, leading to an incorrect direction for the torque vector.

Question 14. Find the torque of the resultant of the three forces represented by \( \overline{F_1} = -3\hat { i } + 6\hat { j } – 3\hat { k } \), \( \overline{F_2} = 4\hat { i } – 10\hat { j } + 12\hat { k } \) and \( \overline{F_3} = 4\hat { i } + 7\hat { j } \) acting at the point with position vector \( \overline{OA} = 8\hat { i } – 6\hat { j } – 4\hat { k } \) about the point with position vector \( \overline{OB} = 18\hat {i } + 3\hat { j } – 9\hat {k} \).
Answer: To find the torque of the resultant force, we first need to sum the individual force vectors to get the total resultant force. Then, we calculate the position vector from the point of rotation to the point of application of force. Finally, we compute their cross product.
**1. Find the resultant force \( \overline{F} \):**
\( \overline{F} = \overline{F_1} + \overline{F_2} + \overline{F_3} \)
\( \overline{F} = (-3\hat { i } + 6\hat { j } – 3\hat { k }) + (4\hat { i } – 10\hat { j } + 12\hat { k }) + (4\hat { i } + 7\hat { j } + 0\hat { k }) \)
\( \implies \overline{F} = (-3+4+4)\hat { i } + (6-10+7)\hat { j } + (-3+12+0)\hat { k } \)
\( \implies \overline{F} = 5\hat { i } + 3\hat { j } + 9\hat { k } \)
**2. Find the position vector \( \vec{r} \):**
Position vector of the point where force acts: \( \overline{OA} = 8\hat { i } – 6\hat { j } – 4\hat { k } \)
Position vector of the point about which torque is calculated: \( \overline{OB} = 18\hat { i } + 3\hat { j } – 9\hat { k } \)
The position vector \( \vec{r} \) (from B to A) is \( \overline{BA} = \overline{OA} - \overline{OB} \):
\( \vec{r} = (8\hat { i } – 6\hat { j } – 4\hat { k }) - (18\hat { i } + 3\hat { j } – 9\hat { k }) \)
\( \implies \vec{r} = (8-18)\hat { i } + (-6-3)\hat { j } + (-4-(-9))\hat { k } \)
\( \implies \vec{r} = -10\hat { i } – 9\hat { j } + 5\hat { k } \)
**3. Calculate the torque \( \vec{\tau} = \vec{r} \times \vec{F} \):**
\[ \vec{\tau} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -10 & -9 & 5 \\ 5 & 3 & 9 \end{array}\right| \]
\( \implies \vec{\tau} = \hat{i}((-9)(9) - (5)(3)) - \hat{j}((-10)(9) - (5)(5)) + \hat{k}((-10)(3) - (-9)(5)) \)
\( \implies \vec{\tau} = \hat{i}(-81 - 15) - \hat{j}(-90 - 25) + \hat{k}(-30 - (-45)) \)
\( \implies \vec{\tau} = \hat{i}(-96) - \hat{j}(-115) + \hat{k}(-30 + 45) \)
\( \implies \vec{\tau} = -96\hat { i } + 115\hat { j } + 15\hat { k } \)
The torque of the resultant force about the given point is \( -96\hat { i } + 115\hat { j } + 15\hat { k } \). This vector represents the rotational effect of all forces combined.
In simple words: First, we added up all the forces to find the single total force. Then, we found the vector from the point of rotation to where this total force acts. Finally, we calculated the cross product of these two vectors to find the twisting force (torque).

🎯 Exam Tip: When multiple forces are involved, calculate their vector sum first to get a single resultant force. Always define the position vector for torque as from the pivot point to the force application point.