RBSE Solutions Class 12 Maths Chapter 13 Vector Exercise 13.2

Question 1. If magnitude of two vectors be 4 and 5 units, then find scalar product of them, where angle between them be:
(i) 60°
(ii) 90°
(iii) 30°
Answer: We know that the scalar product (dot product) of two vectors \( \vec{a} \) and \( \vec{b} \) is given by the formula: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \). Here, \( |\vec{a}| = 4 \) units and \( |\vec{b}| = 5 \) units.
(i) When \( \theta = 60^\circ \):
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 60^\circ \)
\( \vec{a} \cdot \vec{b} = 4 \times 5 \times \frac{1}{2} \)
\( \vec{a} \cdot \vec{b} = 20 \times \frac{1}{2} \)
\( \vec{a} \cdot \vec{b} = 10 \)
The scalar product of the vectors is 10.
(ii) When \( \theta = 90^\circ \):
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 90^\circ \)
\( \vec{a} \cdot \vec{b} = 4 \times 5 \times 0 \)
\( \vec{a} \cdot \vec{b} = 0 \)
The scalar product of the vectors is 0.
(iii) When \( \theta = 30^\circ \):
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 30^\circ \)
\( \vec{a} \cdot \vec{b} = 4 \times 5 \times \frac{\sqrt{3}}{2} \)
\( \vec{a} \cdot \vec{b} = 20 \times \frac{\sqrt{3}}{2} \)
\( \vec{a} \cdot \vec{b} = 10\sqrt{3} \)
The scalar product of the vectors is \( 10\sqrt{3} \).
In simple words: To find the scalar product, multiply the lengths of the two vectors by the cosine of the angle between them. If the angle is 90 degrees, the product is always zero because cos(90°) is zero.

🎯 Exam Tip: Remember the cosine values for common angles (0°, 30°, 45°, 60°, 90°) as they are frequently used in vector problems. A scalar product of zero always implies the vectors are perpendicular.

Question 2. Find \( \overrightarrow {a} . \overrightarrow {b} \), if \( \overrightarrow {a} \) and \( \overrightarrow {b} \) are:
(i) \( \vec{a} = 2\hat{i}+5\hat{j} \) and \( \vec{b} = 3\hat{i}-2\hat{j} \)
(ii) \( \vec{a} = 4\hat{i}+3\hat{k} \) and \( \vec{b} = \hat{i}-\hat{j}+\hat{k} \)
(iii) \( \vec{a} = 5\hat{i}+\hat{j}-2\hat{k} \) and \( \vec{b} = 2\hat{i}-3\hat{j} \)
Answer: To find the scalar product (dot product) of two vectors \( \vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k} \) and \( \vec{b} = b_x\hat{i} + b_y\hat{j} + b_z\hat{k} \), we use the formula: \( \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z \). We multiply the corresponding components and add them up.
(i) Given vectors are \( \vec{a} = 2\hat{i}+5\hat{j} \) and \( \vec{b} = 3\hat{i}-2\hat{j} \). We can write them as \( \vec{a} = 2\hat{i}+5\hat{j}+0\hat{k} \) and \( \vec{b} = 3\hat{i}-2\hat{j}+0\hat{k} \).
Now, we calculate the dot product:
\( \vec{a} \cdot \vec{b} = (2 \times 3) + (5 \times (-2)) + (0 \times 0) \)
\( = 6 - 10 + 0 \)
\( = -4 \)
(ii) Given vectors are \( \vec{a} = 4\hat{i}+3\hat{k} \) and \( \vec{b} = \hat{i}-\hat{j}+\hat{k} \). We can write them as \( \vec{a} = 4\hat{i}+0\hat{j}+3\hat{k} \) and \( \vec{b} = 1\hat{i}-1\hat{j}+1\hat{k} \).
Now, we calculate the dot product:
\( \vec{a} \cdot \vec{b} = (4 \times 1) + (0 \times (-1)) + (3 \times 1) \)
\( = 4 + 0 + 3 \)
\( = 7 \)
(iii) Given vectors are \( \vec{a} = 5\hat{i}+\hat{j}-2\hat{k} \) and \( \vec{b} = 2\hat{i}-3\hat{j} \). We can write them as \( \vec{a} = 5\hat{i}+1\hat{j}-2\hat{k} \) and \( \vec{b} = 2\hat{i}-3\hat{j}+0\hat{k} \).
Now, we calculate the dot product:
\( \vec{a} \cdot \vec{b} = (5 \times 2) + (1 \times (-3)) + ((-2) \times 0) \)
\( = 10 - 3 + 0 \)
\( = 7 \)
In simple words: To find the dot product of two vectors, you just multiply the numbers in front of the \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) parts separately, and then add those three results together. If a part is missing, its number is zero.

🎯 Exam Tip: Be careful with the signs (positive or negative) of the components. Also, remember that if a component (like \( \hat{j} \) in \( 4\hat{i}+3\hat{k} \)) is missing, its coefficient is 0.

Question 4. If coordinates of P and Q are (3, 4) and (12, 4) respectively, then find \( \angle \text{POQ} \) where O is origin.
Answer: Given the coordinates of point P are (3, 4) and point Q are (12, 9). The origin O is (0, 0).
First, we find the position vectors of P and Q with respect to the origin:
\( \vec{OP} = 3\hat{i} + 4\hat{j} \)
\( \vec{OQ} = 12\hat{i} + 9\hat{j} \)
The angle \( \theta \) between two vectors \( \vec{A} \) and \( \vec{B} \) is given by \( \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \). In our case, \( \vec{A} = \vec{OP} \) and \( \vec{B} = \vec{OQ} \).
Calculate the dot product \( \vec{OP} \cdot \vec{OQ} \):
\( \vec{OP} \cdot \vec{OQ} = (3)(12) + (4)(9) \)
\( = 36 + 36 \)
\( = 72 \)
Calculate the magnitudes of \( \vec{OP} \) and \( \vec{OQ} \):
\( |\vec{OP}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)
\( |\vec{OQ}| = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \)
Now, substitute these values into the cosine formula:
\( \cos \theta = \frac{72}{5 \times 15} \)
\( \cos \theta = \frac{72}{75} \)
Simplify the fraction:
\( \cos \theta = \frac{24}{25} \)
Therefore, the angle \( \angle \text{POQ} \) is:
\( \angle \text{POQ} = \cos^{-1} \left(\frac{24}{25}\right) \)
In simple words: To find the angle between two lines from the origin to points P and Q, we use the dot product formula. We multiply the x-components and y-components, add them up, and then divide by the lengths of the two lines multiplied together. Finally, we take the inverse cosine to get the angle.

🎯 Exam Tip: Always correctly identify the position vectors from the origin to the given points. Remember that the magnitude of a vector \( x\hat{i} + y\hat{j} \) is \( \sqrt{x^2+y^2} \).

Question 5. For which value of \( \lambda \), vectors \( \overrightarrow {a} \) and \( \overrightarrow {b} \) are mutually perpendicular:
(i) \( \vec{a} = 2\hat{i}+\lambda\hat{j}+\hat{k} \) and \( \vec{b} = 4\hat{i}-2\hat{j}-2\hat{k} \)
(ii) \( \vec{a} = 2\hat{i}+3\hat{j}+4\hat{k} \) and \( \vec{b} = 3\hat{i}+2\hat{j}-\lambda\hat{k} \)
Answer: Two vectors are mutually perpendicular if their scalar product (dot product) is zero. We use the formula \( \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z = 0 \).
(i) Given vectors are \( \vec{a} = 2\hat{i}+\lambda\hat{j}+\hat{k} \) and \( \vec{b} = 4\hat{i}-2\hat{j}-2\hat{k} \).
Set their dot product to zero:
\( (2)(4) + (\lambda)(-2) + (1)(-2) = 0 \)
\( 8 - 2\lambda - 2 = 0 \)
\( 6 - 2\lambda = 0 \)
\( 2\lambda = 6 \)
\( \lambda = 3 \)
So, the vectors are perpendicular to each other if \( \lambda = 3 \).
(ii) Given vectors are \( \vec{a} = 2\hat{i}+3\hat{j}+4\hat{k} \) and \( \vec{b} = 3\hat{i}+2\hat{j}-\lambda\hat{k} \).
Set their dot product to zero:
\( (2)(3) + (3)(2) + (4)(-\lambda) = 0 \)
\( 6 + 6 - 4\lambda = 0 \)
\( 12 - 4\lambda = 0 \)
\( 4\lambda = 12 \)
\( \lambda = 3 \)
So, the vectors are perpendicular to each other if \( \lambda = 3 \).
In simple words: If two vectors are at a 90-degree angle to each other, their dot product (multiplying corresponding parts and adding them) must be zero. We use this rule to find the unknown value \( \lambda \).

🎯 Exam Tip: The condition \( \vec{a} \cdot \vec{b} = 0 \) is fundamental for perpendicular vectors. Always remember to equate the dot product to zero when solving for an unknown variable in such problems.

Question 6. Find the projection of the vector \( 4\hat{i}-2\hat{j}+\hat{k} \) on the vector \( 3\hat{i}+6\hat{j}-2\hat{k} \).
Answer: Let \( \vec{a} = 4\hat{i}-2\hat{j}+\hat{k} \) and \( \vec{b} = 3\hat{i}+6\hat{j}-2\hat{k} \).
The projection of vector \( \vec{a} \) on vector \( \vec{b} \) is given by the formula: \( \text{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \).
First, calculate the dot product \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (4)(3) + (-2)(6) + (1)(-2) \)
\( = 12 - 12 - 2 \)
\( = -2 \)
Next, calculate the magnitude of vector \( \vec{b} \):
\( |\vec{b}| = \sqrt{3^2 + 6^2 + (-2)^2} \)
\( = \sqrt{9 + 36 + 4} \)
\( = \sqrt{49} \)
\( = 7 \)
Now, substitute these values into the projection formula:
\( \text{Proj}_{\vec{b}} \vec{a} = \frac{-2}{7} \)
The projection is \( -\frac{2}{7} \). Note that projection can be negative, which means the vectors point in generally opposite directions.
In simple words: To find how much of one vector "falls" onto another, we divide their dot product by the length of the vector we are projecting onto. This tells us the scalar component of one vector along the direction of the other.

🎯 Exam Tip: Remember that the projection of \( \vec{a} \) on \( \vec{b} \) is a scalar quantity. If it's a vector projection, you would multiply the scalar projection by the unit vector in the direction of \( \vec{b} \).

Question 7. If \( \vec{a}=2\hat{i}-16\hat{j}+5\hat{k} \) and \( \vec{b}=3\hat{i}+\hat{j}+2\hat{k} \), then find a vector \( \overrightarrow {c} \), so that \( \overrightarrow {a} \), \( \overrightarrow {b} \), \( \overrightarrow {c} \) represents the sides of a right angled triangle.
Answer: Given vectors are \( \vec{a}=2\hat{i}-16\hat{j}+5\hat{k} \) and \( \vec{b}=3\hat{i}+\hat{j}+2\hat{k} \).
For \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) to form the sides of a triangle, their vector sum must be zero, or one vector must be the sum of the other two. Assuming \( \vec{c} \) is the third side such that \( \vec{a} + \vec{b} = \vec{c} \) (or \( \vec{a} + \vec{c} = \vec{b} \) or \( \vec{b} + \vec{c} = \vec{a} \)). Let's test if \( \vec{a} \) and \( \vec{b} \) are perpendicular.
Calculate the dot product \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (2)(3) + (-16)(1) + (5)(2) \)
\( = 6 - 16 + 10 \)
\( = 16 - 16 \)
\( = 0 \)
Since \( \vec{a} \cdot \vec{b} = 0 \), vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular to each other, meaning the angle between them is 90°. This tells us that \( \vec{a} \) and \( \vec{b} \) can be the two shorter sides (legs) of the right-angled triangle. In a right-angled triangle, the sum of the other two sides equals the hypotenuse. Thus, the third side \( \vec{c} \) must be the hypotenuse, represented as the vector sum of \( \vec{a} \) and \( \vec{b} \).
Therefore, \( \vec{c} = \vec{a} + \vec{b} \).
\( \vec{c} = (2\hat{i}-16\hat{j}+5\hat{k}) + (3\hat{i}+\hat{j}+2\hat{k}) \)
Combine the corresponding components:
\( \vec{c} = (2+3)\hat{i} + (-16+1)\hat{j} + (5+2)\hat{k} \)
\( \vec{c} = 5\hat{i}-15\hat{j}+7\hat{k} \)
The diagram below illustrates this vector addition in a right-angled triangle, where \( \vec{a} \) and \( \vec{b} \) are the legs, and \( \vec{c} \) is the hypotenuse.
Therefore, the required vector \( \vec{c} = 5\hat{i}-15\hat{j}+7\hat{k} \).
In simple words: First, we check if the two given vectors are perpendicular by seeing if their dot product is zero. If they are, then they form the sides of a right angle. The third side, which is the hypotenuse, is simply found by adding these two vectors together.

🎯 Exam Tip: The crucial step here is to verify the perpendicularity of \( \vec{a} \) and \( \vec{b} \) using their dot product. If they were not perpendicular, the problem would require a different approach for forming a right-angled triangle.

Question 8. If \( |\overrightarrow {a} + \overrightarrow {b}| = |\overrightarrow {a} - \overrightarrow {b}| \), then prove that \( \overrightarrow {a} \) and \( \overrightarrow {b} \) are mutually perpendicular vectors.
Answer: We are given that \( |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \).
To remove the magnitudes, we can square both sides of the equation:
\( |\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2 \)
We know that for any vector \( \vec{x} \), \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \). So, we can rewrite the equation as:
\( (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) \)
Now, expand both sides using the distributive property of dot product:
\( \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} \)
Since the dot product is commutative (\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \)) and \( \vec{x} \cdot \vec{x} = |\vec{x}|^2 \), the equation becomes:
\( |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \)
Subtract \( |\vec{a}|^2 \) and \( |\vec{b}|^2 \) from both sides:
\( 2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b}) \)
Add \( 2(\vec{a} \cdot \vec{b}) \) to both sides:
\( 2(\vec{a} \cdot \vec{b}) + 2(\vec{a} \cdot \vec{b}) = 0 \)
\( 4(\vec{a} \cdot \vec{b}) = 0 \)
Divide by 4:
\( \vec{a} \cdot \vec{b} = 0 \)
Since the dot product of \( \vec{a} \) and \( \vec{b} \) is 0, this proves that vectors \( \vec{a} \) and \( \vec{b} \) are mutually perpendicular. This identity shows a special relationship where the diagonals of a parallelogram formed by \( \vec{a} \) and \( \vec{b} \) have equal length only if the sides are perpendicular.
In simple words: When the lengths of the sum and difference of two vectors are equal, it means that the vectors themselves must be perpendicular to each other. We prove this by squaring both sides of the given equation and simplifying using dot product rules until we find their dot product is zero.

🎯 Exam Tip: This is a standard proof. Remember the identity \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \) and the commutative property \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \) to simplify the dot products effectively.

Question 9. If coordinates of points A, B, C and D are (3, 2, 4), (4, 5, -1), (6, 3, 2) and (2, 1, 0) respectively, then prove that lines \( \overrightarrow {AB} \) and \( \overrightarrow {CD} \) are mutually perpendicular.
Answer: Given the coordinates of points A, B, C, and D:
A = (3, 2, 4)
B = (4, 5, -1)
C = (6, 3, 2)
D = (2, 1, 0)
To prove that lines \( \overrightarrow {AB} \) and \( \overrightarrow {CD} \) are mutually perpendicular, we need to show that their dot product is zero. First, we find the position vectors of A, B, C, and D with respect to the origin (O):
\( \vec{OA} = 3\hat{i} + 2\hat{j} + 4\hat{k} \)
\( \vec{OB} = 4\hat{i} + 5\hat{j} - \hat{k} \)
\( \vec{OC} = 6\hat{i} + 3\hat{j} + 2\hat{k} \)
\( \vec{OD} = 2\hat{i} + \hat{j} + 0\hat{k} \)
Now, we find the vectors \( \overrightarrow {AB} \) and \( \overrightarrow {CD} \):
\( \overrightarrow {AB} = \vec{OB} - \vec{OA} \)
\( = (4\hat{i} + 5\hat{j} - \hat{k}) - (3\hat{i} + 2\hat{j} + 4\hat{k}) \)
\( = (4-3)\hat{i} + (5-2)\hat{j} + (-1-4)\hat{k} \)
\( = \hat{i} + 3\hat{j} - 5\hat{k} \)
\( \overrightarrow {CD} = \vec{OD} - \vec{OC} \)
\( = (2\hat{i} + \hat{j} + 0\hat{k}) - (6\hat{i} + 3\hat{j} + 2\hat{k}) \)
\( = (2-6)\hat{i} + (1-3)\hat{j} + (0-2)\hat{k} \)
\( = -4\hat{i} - 2\hat{j} - 2\hat{k} \)
Finally, calculate the dot product of \( \overrightarrow {AB} \) and \( \overrightarrow {CD} \):
\( \overrightarrow {AB} \cdot \overrightarrow {CD} = (\hat{i} + 3\hat{j} - 5\hat{k}) \cdot (-4\hat{i} - 2\hat{j} - 2\hat{k}) \)
\( = (1)(-4) + (3)(-2) + (-5)(-2) \)
\( = -4 - 6 + 10 \)
\( = -10 + 10 \)
\( = 0 \)
Since \( \overrightarrow {AB} \cdot \overrightarrow {CD} = 0 \), the vectors \( \overrightarrow {AB} \) and \( \overrightarrow {CD} \) are mutually perpendicular. Thus, the lines AB and CD are perpendicular to each other. This shows that the angle formed by these two line segments is 90 degrees.
In simple words: To show that two lines are perpendicular, we first find the vectors representing those lines by subtracting their start and end point coordinates. Then, we calculate the dot product of these two vectors. If the dot product is zero, the lines are perpendicular.

🎯 Exam Tip: Remember that a vector from point P to point Q is given by \( \vec{OQ} - \vec{OP} \). Always be careful with signs when subtracting coordinates.

Question 10. For any vector \( \overrightarrow {a} \), prove that \( \vec{a} = (\vec{a} \cdot \hat{i})\hat{i}+(\vec{a} \cdot \hat{j})\hat{j}+(\vec{a} \cdot \hat{k})\hat{k} \).
Answer: Let \( \vec{a} \) be any vector in 3D space. We can express it in terms of its components along the x, y, and z axes:
\( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \)
Now, let's calculate the dot product of \( \vec{a} \) with each of the unit vectors \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \). We know the properties of unit vectors: \( \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 \) and \( \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0 \).
1. Dot product with \( \hat{i} \):
\( \vec{a} \cdot \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot \hat{i} \)
\( = a_1(\hat{i} \cdot \hat{i}) + a_2(\hat{j} \cdot \hat{i}) + a_3(\hat{k} \cdot \hat{i}) \)
\( = a_1(1) + a_2(0) + a_3(0) \)
\( = a_1 \)
So, \( a_1 = \vec{a} \cdot \hat{i} \).
2. Dot product with \( \hat{j} \):
\( \vec{a} \cdot \hat{j} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot \hat{j} \)
\( = a_1(\hat{i} \cdot \hat{j}) + a_2(\hat{j} \cdot \hat{j}) + a_3(\hat{k} \cdot \hat{j}) \)
\( = a_1(0) + a_2(1) + a_3(0) \)
\( = a_2 \)
So, \( a_2 = \vec{a} \cdot \hat{j} \).
3. Dot product with \( \hat{k} \):
\( \vec{a} \cdot \hat{k} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot \hat{k} \)
\( = a_1(\hat{i} \cdot \hat{k}) + a_2(\hat{j} \cdot \hat{k}) + a_3(\hat{k} \cdot \hat{k}) \)
\( = a_1(0) + a_2(0) + a_3(1) \)
\( = a_3 \)
So, \( a_3 = \vec{a} \cdot \hat{k} \).
Now, substitute the values of \( a_1, a_2, \) and \( a_3 \) back into the original expression for \( \vec{a} \):
\( \vec{a} = (\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k} \)
This fundamental identity shows how any vector can be completely represented by its components projected onto the coordinate axes. It is a cornerstone of vector analysis.
In simple words: This proof shows that you can write any vector by adding up its "shadows" (projections) on the x, y, and z axes. Each shadow is found by taking the dot product of the vector with the unit vector along that axis, then multiplying by that same unit vector.

🎯 Exam Tip: The key to this proof is understanding the properties of orthonormal unit vectors (\( \hat{i} \cdot \hat{i} = 1 \) and \( \hat{i} \cdot \hat{j} = 0 \)). Clearly show each step of the dot product calculation.

Question 11. Use vectors to prove that sum of square of diagonal of a parallelogram is equal to the sum of square of their side.
Answer: Let OACB be a parallelogram. Let O be the origin. The position vectors of points A and B are \( \vec{OA} = \vec{a} \) and \( \vec{OB} = \vec{b} \), respectively.
In a parallelogram, opposite sides are equal and parallel. So, \( \overrightarrow{OC} \) is one diagonal and \( \overrightarrow{AB} \) is the other diagonal.
From vector addition in triangle OAC:
\( \vec{OC} = \vec{OA} + \vec{AC} \)
Since OACB is a parallelogram, \( \vec{AC} = \vec{OB} = \vec{b} \).
So, the first diagonal is \( \vec{OC} = \vec{a} + \vec{b} \).
The other diagonal is \( \overrightarrow{AB} \). From triangle OAB:
\( \overrightarrow{AB} = \vec{OB} - \vec{OA} \)
So, the second diagonal is \( \overrightarrow{AB} = \vec{b} - \vec{a} \).
The sides of the parallelogram are \( |\vec{OA}| = |\vec{a}| \) and \( |\vec{OB}| = |\vec{b}| \).
Now, we need to prove: \( |\vec{OC}|^2 + |\overrightarrow{AB}|^2 = 2(|\vec{OA}|^2 + |\vec{OB}|^2) \)
Substitute the vector expressions for the diagonals:
\( |\vec{a} + \vec{b}|^2 + |\vec{b} - \vec{a}|^2 \)
Using the property \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \):
\( (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) + (\vec{b} - \vec{a}) \cdot (\vec{b} - \vec{a}) \)
Expand the dot products:
\( (\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}) + (\vec{b} \cdot \vec{b} - \vec{b} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{a}) \)
Since \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \), this simplifies to:
\( (|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2) + (|\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{a}|^2) \)
Now, combine like terms:
\( |\vec{a}|^2 + |\vec{b}|^2 + |\vec{b}|^2 + |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) - 2(\vec{a} \cdot \vec{b}) \)
\( = 2|\vec{a}|^2 + 2|\vec{b}|^2 \)
\( = 2(|\vec{a}|^2 + |\vec{b}|^2) \)
This matches \( 2(|\vec{OA}|^2 + |\vec{OB}|^2) \).
Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. This is a vector form of the parallelogram law, a geometric identity.
In simple words: We prove that if you square the lengths of both diagonals of a parallelogram and add them up, this total will be the same as taking the lengths of its two different sides, squaring them, adding them, and then multiplying that sum by two. We use vector addition and dot product properties to show this.

🎯 Exam Tip: Clearly define the position vectors for the vertices and the diagonals. The use of \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \) and the expansion of \( (\vec{a} \pm \vec{b}) \cdot (\vec{a} \pm \vec{b}) \) are central to this proof.

Question 1. If magnitude of two vectors be 4 and 5 units, then find scalar product of them, where angle between them be:
(i) 60°
(ii) 90°
(iii) 30°
Answer:
(i) Given: \( \theta = 60^\circ \), \( |\vec{a}| = 4 \), \( |\vec{b}| = 5 \)
We know that the scalar product is given by \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \).
So, we have:
\( \cos 60^\circ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \)

\( \implies \frac{1}{2} = \frac{\vec{a} \cdot \vec{b}}{4 \times 5} \)

\( \implies \frac{1}{2} = \frac{\vec{a} \cdot \vec{b}}{20} \)

\( \implies \vec{a} \cdot \vec{b} = \frac{20}{2} \)

\( \implies \vec{a} \cdot \vec{b} = 10 \)
Therefore, the scalar product of the vectors is 10.
(ii) Given: \( \theta = 90^\circ \), \( |\vec{a}| = 4 \), \( |\vec{b}| = 5 \)
We use the same formula: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \).
So, we have:
\( \cos 90^\circ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \)

\( \implies 0 = \frac{\vec{a} \cdot \vec{b}}{4 \times 5} \)

\( \implies 0 = \frac{\vec{a} \cdot \vec{b}}{20} \)

\( \implies \vec{a} \cdot \vec{b} = 0 \)
Therefore, the scalar product of the vectors is 0. This makes sense because perpendicular vectors have a dot product of zero.
(iii) Given: \( \theta = 30^\circ \), \( |\vec{a}| = 4 \), \( |\vec{b}| = 5 \)
Using the formula: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \).
So, we have:
\( \cos 30^\circ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \)

\( \implies \frac{\sqrt{3}}{2} = \frac{\vec{a} \cdot \vec{b}}{4 \times 5} \)

\( \implies \frac{\sqrt{3}}{2} = \frac{\vec{a} \cdot \vec{b}}{20} \)

\( \implies \vec{a} \cdot \vec{b} = \frac{20\sqrt{3}}{2} \)

\( \implies \vec{a} \cdot \vec{b} = 10\sqrt{3} \)
Therefore, the scalar product of the vectors is \( 10\sqrt{3} \).
In simple words: To find the scalar product of two vectors, multiply their lengths and then multiply by the cosine of the angle between them. If the angle is 90 degrees, the product is zero.

🎯 Exam Tip: Remember the formula for scalar product \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \) and recall common cosine values for 0°, 30°, 60°, and 90°.

Question 2. Find \( \vec{a} \cdot \vec{b} \), if \( \vec{a} \) and \( \vec{b} \) are as:
(i) \( \vec{a} = 2\hat{i}+5\hat{j} \) and \( \vec{b} = 3\hat{i}-2\hat{j} \)
(ii) \( \vec{a} = 4\hat{i}+3\hat{k} \) and \( \vec{b} = \hat{i}-\hat{j}+\hat{k} \)
(iii) \( \vec{a} = 5\hat{i}+\hat{j}-2\hat{k} \) and \( \vec{b} = 2\hat{i}-3\hat{j} \)
Answer:
(i) Given vectors are \( \vec{a} = 2\hat{i}+5\hat{j} \) and \( \vec{b} = 3\hat{i}-2\hat{j} \).
We can write them as \( \vec{a} = 2\hat{i}+5\hat{j}+0\hat{k} \) and \( \vec{b} = 3\hat{i}-2\hat{j}+0\hat{k} \).
To find the scalar product, we multiply the corresponding components and add them up:
\( \vec{a} \cdot \vec{b} = (2\hat{i}+5\hat{j}+0\hat{k}) \cdot (3\hat{i}-2\hat{j}+0\hat{k}) \)
\( = (2 \times 3) + (5 \times (-2)) + (0 \times 0) \)
\( = 6 - 10 + 0 \)
\( = -4 \)
So, the scalar product \( \vec{a} \cdot \vec{b} = -4 \).
(ii) Given vectors are \( \vec{a} = 4\hat{i}+3\hat{k} \) and \( \vec{b} = \hat{i}-\hat{j}+\hat{k} \).
We can write \( \vec{a} = 4\hat{i}+0\hat{j}+3\hat{k} \).
Now, we find the scalar product:
\( \vec{a} \cdot \vec{b} = (4\hat{i}+0\hat{j}+3\hat{k}) \cdot (\hat{i}-\hat{j}+\hat{k}) \)
\( = (4 \times 1) + (0 \times (-1)) + (3 \times 1) \)
\( = 4 + 0 + 3 \)
\( = 7 \)
So, the scalar product \( \vec{a} \cdot \vec{b} = 7 \).
(iii) Given vectors are \( \vec{a} = 5\hat{i}+\hat{j}-2\hat{k} \) and \( \vec{b} = 2\hat{i}-3\hat{j} \).
We can write \( \vec{b} = 2\hat{i}-3\hat{j}+0\hat{k} \).
Now, we find the scalar product:
\( \vec{a} \cdot \vec{b} = (5\hat{i}+\hat{j}-2\hat{k}) \cdot (2\hat{i}-3\hat{j}+0\hat{k}) \)
\( = (5 \times 2) + (1 \times (-3)) + ((-2) \times 0) \)
\( = 10 - 3 + 0 \)
\( = 7 \)
So, the scalar product \( \vec{a} \cdot \vec{b} = 7 \). The scalar product helps understand the geometric relationship between vectors, like whether they are perpendicular.
In simple words: To find the dot product of two vectors, multiply their matching parts (i, j, and k components) and then add those results together. This gives a single number.

🎯 Exam Tip: Ensure you correctly identify the corresponding components (i with i, j with j, k with k) when calculating the scalar product. If a component is missing, assume its coefficient is zero.

Question 4. If coordinates of P and Q are (3, 4) and (12, 4) respectively, then find \( \angle POQ \) where O is origin.
Answer:
Given coordinates of P are (3, 4) and Q are (12, 4). O is the origin (0, 0).
The position vector \( \vec{OP} \) from the origin to P is \( 3\hat{i}+4\hat{j} \).
The position vector \( \vec{OQ} \) from the origin to Q is \( 12\hat{i}+9\hat{j} \).
Let \( \angle POQ = \theta \). We can find this angle using the dot product formula:
\( \cos \theta = \frac{\vec{OP} \cdot \vec{OQ}}{|\vec{OP}| |\vec{OQ}|} \)
First, calculate the dot product \( \vec{OP} \cdot \vec{OQ} \):
\( \vec{OP} \cdot \vec{OQ} = (3\hat{i}+4\hat{j}) \cdot (12\hat{i}+9\hat{j}) \)
\( = (3 \times 12) + (4 \times 9) \)
\( = 36 + 36 \)
\( = 72 \)
Next, calculate the magnitudes \( |\vec{OP}| \) and \( |\vec{OQ}| \):
\( |\vec{OP}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)
\( |\vec{OQ}| = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \)
Now, substitute these values into the cosine formula:
\( \cos \theta = \frac{72}{5 \times 15} \)
\( = \frac{72}{75} \)
Simplify the fraction:
\( \cos \theta = \frac{24}{25} \)
To find the angle \( \theta \), we take the inverse cosine:
\( \theta = \cos^{-1}\left(\frac{24}{25}\right) \)
Thus, the angle \( \angle POQ \) is \( \cos^{-1}\left(\frac{24}{25}\right) \). This method uses vector properties to find angles in a coordinate plane.
In simple words: To find the angle between two lines from the origin, use the dot product of their position vectors. Divide the dot product by the product of their lengths, then take the inverse cosine of that number to get the angle.

🎯 Exam Tip: Remember that position vectors from the origin (0,0) to a point (x,y) are simply \( x\hat{i}+y\hat{j} \). Be careful with calculations of dot products and magnitudes.

Question 5. For which value of \( \lambda \), vectors \( \vec{a} \) and \( \vec{b} \) are mutually perpendicular :
(i) \( \vec{a} = 2\hat{i}+\lambda\hat{j}+\hat{k}; \vec{b} = 4\hat{i}-2\hat{j}-2\hat{k} \)
(ii) \( \vec{a} = 2\hat{i}+3\hat{j}+4\hat{k} \) and \( \vec{b} = 3\hat{i}+2\hat{j}-\lambda\hat{k} \)
Answer:
Two vectors are mutually perpendicular if their scalar (dot) product is zero.
(i) Given vectors: \( \vec{a} = 2\hat{i}+\lambda\hat{j}+\hat{k} \) and \( \vec{b} = 4\hat{i}-2\hat{j}-2\hat{k} \).
Since they are perpendicular, their dot product must be 0:
\( \vec{a} \cdot \vec{b} = 0 \)
\( (2\hat{i}+\lambda\hat{j}+\hat{k}) \cdot (4\hat{i}-2\hat{j}-2\hat{k}) = 0 \)
\( (2 \times 4) + (\lambda \times (-2)) + (1 \times (-2)) = 0 \)
\( 8 - 2\lambda - 2 = 0 \)
\( 6 - 2\lambda = 0 \)

\( \implies 2\lambda = 6 \)

\( \implies \lambda = 3 \)
Hence, the vectors are perpendicular to each other if \( \lambda = 3 \).
(ii) Given vectors: \( \vec{a} = 2\hat{i}+3\hat{j}+4\hat{k} \) and \( \vec{b} = 3\hat{i}+2\hat{j}-\lambda\hat{k} \).
Since they are perpendicular, their dot product must be 0:
\( \vec{a} \cdot \vec{b} = 0 \)
\( (2\hat{i}+3\hat{j}+4\hat{k}) \cdot (3\hat{i}+2\hat{j}-\lambda\hat{k}) = 0 \)
\( (2 \times 3) + (3 \times 2) + (4 \times (-\lambda)) = 0 \)
\( 6 + 6 - 4\lambda = 0 \)
\( 12 - 4\lambda = 0 \)

\( \implies 4\lambda = 12 \)

\( \implies \lambda = 3 \)
Hence, the vectors are perpendicular to each other if \( \lambda = 3 \). Finding this value helps in understanding spatial relationships of vectors.
In simple words: If two vectors are at a right angle (perpendicular), their dot product is always zero. Use this rule to find the unknown value that makes them perpendicular.

🎯 Exam Tip: The condition for perpendicular vectors is \( \vec{a} \cdot \vec{b} = 0 \). This is a fundamental concept in vector algebra and is frequently tested.

Question 6. Find the projection of the vector \( 4\hat{i}-2\hat{j}+\hat{k} \) on the vector \( 3\hat{i}+6\hat{j}-2\hat{k} \)
Answer:
Let \( \vec{a} = 4\hat{i}-2\hat{j}+\hat{k} \) and \( \vec{b} = 3\hat{i}+6\hat{j}-2\hat{k} \).
The projection of vector \( \vec{a} \) on vector \( \vec{b} \) is given by the formula:
\( \text{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \)
First, calculate the dot product \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (4\hat{i}-2\hat{j}+\hat{k}) \cdot (3\hat{i}+6\hat{j}-2\hat{k}) \)
\( = (4 \times 3) + ((-2) \times 6) + (1 \times (-2)) \)
\( = 12 - 12 - 2 \)
\( = -2 \)
Next, calculate the magnitude of vector \( \vec{b} \):
\( |\vec{b}| = \sqrt{3^2 + 6^2 + (-2)^2} \)
\( = \sqrt{9 + 36 + 4} \)
\( = \sqrt{49} \)
\( = 7 \)
Now, substitute these values into the projection formula:
\( \text{Proj}_{\vec{b}} \vec{a} = \frac{-2}{7} \)
The projection is \( -\frac{2}{7} \). Although the result is negative, projection itself is a scalar quantity and represents the component of one vector along the direction of another. The negative sign indicates that the two vectors generally point in opposite directions relative to each other.
In simple words: To find how much of one vector "points in the same direction" as another, calculate their dot product and divide by the length of the vector it's being projected onto.

🎯 Exam Tip: The projection of vector \( \vec{a} \) on \( \vec{b} \) is a scalar value \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \). Remember that projection can be negative if the angle between the vectors is obtuse (greater than 90°).

Question 7. If \( \vec{a}=2\hat{i}-16\hat{j}+5\hat{k} \) and \( \vec{b}=3\hat{i}+\hat{j}+2\hat{k} \) then find a vector \( \vec{c} \), so that \( \vec{a} \), \( \vec{b} \), \( \vec{c} \) represents the sides of a right angled triangle.
Answer:
Given vectors are \( \vec{a}=2\hat{i}-16\hat{j}+5\hat{k} \) and \( \vec{b}=3\hat{i}+\hat{j}+2\hat{k} \).
For a right-angled triangle, the sum of two side vectors can give the third side vector (triangle law of vector addition). Also, for vectors to form a right-angled triangle, two of the vectors must be perpendicular, meaning their dot product is zero.
Let's check if \( \vec{a} \) and \( \vec{b} \) are perpendicular:
\( \vec{a} \cdot \vec{b} = (2\hat{i}-16\hat{j}+5\hat{k}) \cdot (3\hat{i}+\hat{j}+2\hat{k}) \)
\( = (2 \times 3) + ((-16) \times 1) + (5 \times 2) \)
\( = 6 - 16 + 10 \)
\( = 0 \)
Since \( \vec{a} \cdot \vec{b} = 0 \), vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular to each other. This means they can form the two shorter sides (legs) of a right-angled triangle.
According to the triangle law of vector addition, if \( \vec{a} \) and \( \vec{b} \) are two sides, the third side \( \vec{c} \) (which would be the hypotenuse if \( \vec{a} \) and \( \vec{b} \) are the legs) is given by \( \vec{c} = \vec{a} + \vec{b} \).
So, let's find \( \vec{c} \):
\( \vec{c} = (2\hat{i}-16\hat{j}+5\hat{k}) + (3\hat{i}+\hat{j}+2\hat{k}) \)
\( = (2+3)\hat{i} + (-16+1)\hat{j} + (5+2)\hat{k} \)
\( = 5\hat{i} - 15\hat{j} + 7\hat{k} \)
Thus, the required vector \( \vec{c} \) that forms a right-angled triangle with \( \vec{a} \) and \( \vec{b} \) is \( 5\hat{i} - 15\hat{j} + 7\hat{k} \). This calculation confirms the conditions for forming a right-angled triangle with vectors.
In simple words: If two vectors are perpendicular, they can be the shorter sides of a right-angled triangle. The third side, the longest one (hypotenuse), is found by adding these two vectors together.

🎯 Exam Tip: For vectors to form a right-angled triangle, two of the vectors must be perpendicular (their dot product is zero), and the third vector will be the sum of the other two, representing the hypotenuse.

Question 8. If \( | \vec{a} + \vec{b} | = | \vec{a} - \vec{b} | \), then prove that \( \vec{a} \) and \( \vec{b} \) are mutually perpendicular vectors.
Answer:
Given that \( | \vec{a} + \vec{b} | = | \vec{a} - \vec{b} | \).
We know that \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \). So, we can square both sides of the given equation:
\( | \vec{a} + \vec{b} |^2 = | \vec{a} - \vec{b} |^2 \)
\( (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) \)
Using the distributive law for dot products:
\( \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} \)
Since \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 \), \( \vec{b} \cdot \vec{b} = |\vec{b}|^2 \), and the dot product is commutative (\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \)):
\( |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \)
Subtract \( |\vec{a}|^2 \) and \( |\vec{b}|^2 \) from both sides:
\( 2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b}) \)
Add \( 2(\vec{a} \cdot \vec{b}) \) to both sides:
\( 4(\vec{a} \cdot \vec{b}) = 0 \)

\( \implies \vec{a} \cdot \vec{b} = 0 \)
Since the dot product \( \vec{a} \cdot \vec{b} \) is 0, the vectors \( \vec{a} \) and \( \vec{b} \) must be mutually perpendicular. This property is useful in geometry and physics to determine angles between forces or displacements.
In simple words: If the length of adding two vectors is the same as the length of subtracting them, it means the vectors must be at a 90-degree angle to each other.

🎯 Exam Tip: This is a standard proof. The key step is to square both sides of the given equality and then expand the dot products using the property \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \).

Question 9. If coordinates of points A, B, C and D are (3, 2, 4), (4, 5, -1), (6, 3, 2) and (2, 1, 0) respectively, then prove that lines \( \vec{AB} \) and \( \vec{CD} \) are mutually perpendicular.
Answer:
Given coordinates of points: A(3, 2, 4), B(4, 5, -1), C(6, 3, 2), and D(2, 1, 0).
First, we find the position vectors of these points from the origin O:
\( \vec{OA} = 3\hat{i}+2\hat{j}+4\hat{k} \)
\( \vec{OB} = 4\hat{i}+5\hat{j}-1\hat{k} \)
\( \vec{OC} = 6\hat{i}+3\hat{j}+2\hat{k} \)
\( \vec{OD} = 2\hat{i}+1\hat{j}+0\hat{k} \)
Now, we find the vectors \( \vec{AB} \) and \( \vec{CD} \):
\( \vec{AB} = \vec{OB} - \vec{OA} \)
\( = (4\hat{i}+5\hat{j}-1\hat{k}) - (3\hat{i}+2\hat{j}+4\hat{k}) \)
\( = (4-3)\hat{i} + (5-2)\hat{j} + (-1-4)\hat{k} \)
\( = \hat{i} + 3\hat{j} - 5\hat{k} \)
And,
\( \vec{CD} = \vec{OD} - \vec{OC} \)
\( = (2\hat{i}+\hat{j}+0\hat{k}) - (6\hat{i}+3\hat{j}+2\hat{k}) \)
\( = (2-6)\hat{i} + (1-3)\hat{j} + (0-2)\hat{k} \)
\( = -4\hat{i} - 2\hat{j} - 2\hat{k} \)
To prove that \( \vec{AB} \) and \( \vec{CD} \) are mutually perpendicular, we need to show their dot product is zero:
\( \vec{AB} \cdot \vec{CD} = (\hat{i}+3\hat{j}-5\hat{k}) \cdot (-4\hat{i}-2\hat{j}-2\hat{k}) \)
\( = (1 \times (-4)) + (3 \times (-2)) + ((-5) \times (-2)) \)
\( = -4 - 6 + 10 \)
\( = -10 + 10 \)
\( = 0 \)
Since \( \vec{AB} \cdot \vec{CD} = 0 \), vectors \( \vec{AB} \) and \( \vec{CD} \) are mutually perpendicular. This means the lines formed by these vectors cross each other at a right angle.
In simple words: First, find the vectors for the lines AB and CD. Then, calculate their dot product. If the answer is zero, it proves that the lines are perpendicular.

🎯 Exam Tip: Remember that a vector between two points A and B is given by \( \vec{OB} - \vec{OA} \). The condition for perpendicularity is always a zero dot product between the two vectors.

Question 10. For any vector \( \vec{a} \), prove that \( \vec{a} = (\vec{a}.\hat{i})\hat{i}+(\vec{a}.\hat{j})\hat{j}+(\vec{a}.\hat{k})\hat{k} \).
Answer:
Let \( \vec{a} \) be any vector. We can write \( \vec{a} \) in terms of its components along the x, y, and z axes as:
\( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \)
Now, let's find the dot product of \( \vec{a} \) with each of the unit vectors \( \hat{i}, \hat{j}, \hat{k} \):
\( \vec{a} \cdot \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot \hat{i} \)
Since \( \hat{i} \cdot \hat{i} = 1 \) and \( \hat{j} \cdot \hat{i} = 0 \), \( \hat{k} \cdot \hat{i} = 0 \):
\( \vec{a} \cdot \hat{i} = a_1(1) + a_2(0) + a_3(0) = a_1 \)
Similarly, for \( \hat{j} \):
\( \vec{a} \cdot \hat{j} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot \hat{j} \)
\( = a_1(0) + a_2(1) + a_3(0) = a_2 \)
And for \( \hat{k} \):
\( \vec{a} \cdot \hat{k} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot \hat{k} \)
\( = a_1(0) + a_2(0) + a_3(1) = a_3 \)
So, we have found that:
\( a_1 = \vec{a} \cdot \hat{i} \)
\( a_2 = \vec{a} \cdot \hat{j} \)
\( a_3 = \vec{a} \cdot \hat{k} \)
Now, substitute these values back into the original expression for \( \vec{a} \):
\( \vec{a} = (\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k} \)
This proves the identity. This shows how any vector can be broken down into its components along the coordinate axes using dot products.
In simple words: Any vector can be written by adding up its pieces along the x, y, and z directions. Each piece is found by taking the dot product of the vector with the unit vector for that direction, and then multiplying by that unit vector again.

🎯 Exam Tip: This identity is crucial for understanding vector components and expressing a vector in different coordinate systems. Remember the properties of dot products with orthogonal unit vectors (e.g., \( \hat{i} \cdot \hat{i} = 1 \), \( \hat{i} \cdot \hat{j} = 0 \)).

Question 11. Use vectors to prove that sum of square of diagonal of a parallelogram is equal to the sum of square of their side.
Answer:
Let OACB be a parallelogram. Let O be the origin.
Let \( \vec{OA} = \vec{a} \) and \( \vec{OB} = \vec{b} \) be the position vectors of vertices A and B respectively.
In a parallelogram, opposite sides are equal and parallel. So, \( \vec{OC} \) is one diagonal and \( \vec{AB} \) is the other diagonal.
Using the triangle law of vector addition:
The diagonal \( \vec{OC} = \vec{OA} + \vec{AC} \). Since \( \vec{AC} = \vec{OB} \), we have \( \vec{OC} = \vec{a} + \vec{b} \).
The other diagonal \( \vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a} \).
We need to prove that \( |\vec{OC}|^2 + |\vec{AB}|^2 = |\vec{OA}|^2 + |\vec{AC}|^2 + |\vec{CB}|^2 + |\vec{OB}|^2 \).
Since \( |\vec{AC}| = |\vec{OB}| = |\vec{b}| \) and \( |\vec{CB}| = |\vec{OA}| = |\vec{a}| \), the right side simplifies to \( |\vec{a}|^2 + |\vec{b}|^2 + |\vec{a}|^2 + |\vec{b}|^2 = 2|\vec{a}|^2 + 2|\vec{b}|^2 \).
So, we need to prove: \( |\vec{OC}|^2 + |\vec{AB}|^2 = 2|\vec{a}|^2 + 2|\vec{b}|^2 \).
Let's calculate \( |\vec{OC}|^2 \):
\( |\vec{OC}|^2 = |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) \)
\( = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} \)
\( = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \)
Next, calculate \( |\vec{AB}|^2 \):
\( |\vec{AB}|^2 = |\vec{b} - \vec{a}|^2 = (\vec{b} - \vec{a}) \cdot (\vec{b} - \vec{a}) \)
\( = \vec{b} \cdot \vec{b} - \vec{b} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{a} \)
\( = |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{a}|^2 \)
Now, sum the squares of the diagonals:
\( |\vec{OC}|^2 + |\vec{AB}|^2 = (|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2) + (|\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2) \)
\( = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{a}|^2 + |\vec{b}|^2 \)
\( = 2|\vec{a}|^2 + 2|\vec{b}|^2 \)
Since \( |\vec{a}| = |\vec{OA}| = |\vec{CB}| \) and \( |\vec{b}| = |\vec{OB}| = |\vec{AC}| \), the sum of the squares of the diagonals is equal to twice the sum of the squares of the adjacent sides (or the sum of the squares of all four sides). This identity is a fundamental property of parallelograms in geometry.
In simple words: If you add the squares of the lengths of the two diagonals of a parallelogram, that total will be equal to twice the sum of the squares of the lengths of its two different sides.

🎯 Exam Tip: This proof relies on defining the diagonals as vector sums and differences, then using the property \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \) and the distributive law of dot products.