OP Malhotra Class 12 Maths Solutions Chapter 21 Vectors Exercise 21 (C)

Question 1. If the position vector of a point (-4, -3) be \( \vec{a} \), find \( |\vec{a}| \).
Answer:
Given \( \vec{a} = -4 \hat{i} - 3 \hat{j} \)
The magnitude of a vector is calculated using the square root of the sum of the squares of its components.
\( \implies |\vec{a}| = \sqrt{(-4)^2+(-3)^2} \)
\( = \sqrt{16+9} \)
\( = \sqrt{25} \)
\( = 5 \)
In simple words: We are given a vector that shows the position of a point. To find its length (magnitude), we take the square root of the sum of the squares of its x and y parts. This gives us the final length of the vector.

🎯 Exam Tip: Remember that the magnitude of a vector \( \vec{v} = x\hat{i} + y\hat{j} \) is always a positive scalar value, calculated as \( \sqrt{x^2+y^2} \). The signs of x and y do not affect the final magnitude because they are squared.

Question 2. If P(-1, 3) and Q(2, -7) express the vector \( \vec{r} = \overrightarrow{\mathrm{PQ}} \) in terms of unit vectors \( \hat{i} \) and \( \hat{j} \). Also determine the magnitude of \( \vec{r} \).
Answer:
Given P.V of P \( = -\hat{i} + 3 \hat{j} \)
P.V of Q \( = 2 \hat{i} - 7 \hat{j} \)
To find the vector from P to Q, we subtract the position vector of P from the position vector of Q.
\( \implies \vec{r} = \overrightarrow{\mathrm{PQ}} = \text{P.V of Q} - \text{P.V of P} \)
\( = (2 \hat{i} - 7 \hat{j}) - (-\hat{i} + 3 \hat{j}) \)
\( = 2 \hat{i} - 7 \hat{j} + \hat{i} - 3 \hat{j} \)
\( = 3 \hat{i} - 10 \hat{j} \)
Now, we find the magnitude of this vector.
\( \implies |\vec{r}| = \sqrt{3^2+(-10)^2} \)
\( = \sqrt{9+100} \)
\( = \sqrt{109} \)
In simple words: First, we find the vector from point P to point Q by subtracting their position vectors. Then, we find the length of this new vector using the Pythagorean theorem with its x and y components.

🎯 Exam Tip: When finding a vector between two points, \( \overrightarrow{AB} \), always remember it's "position vector of B minus position vector of A" (terminal minus initial). Be careful with signs during subtraction.

Question 3. If \( \vec{a} = 3 \hat{i} - 2 \hat{j} \) and \( \vec{b} = -\hat{i} + \hat{j} \), find the magnitude of \( 2 \vec{a} - 3 \vec{b} \).
Answer:
Given \( \vec{a} = 3 \hat{i} - 2 \hat{j} \) and \( \vec{b} = -\hat{i} + \hat{j} \)
First, we calculate the new vector \( 2 \vec{a} - 3 \vec{b} \). This involves scalar multiplication and vector subtraction.
\( 2 \vec{a} - 3 \vec{b} = 2(3 \hat{i} - 2 \hat{j}) - 3(-\hat{i} + \hat{j}) \)
\( = 6 \hat{i} - 4 \hat{j} + 3 \hat{i} - 3 \hat{j} \)
\( = 9 \hat{i} - 7 \hat{j} \)
Now, we find the magnitude of this resulting vector.
\( \implies |2 \vec{a} - 3 \vec{b}| = |9 \hat{i} - 7 \hat{j}| \)
\( = \sqrt{9^2+(-7)^2} \)
\( = \sqrt{81+49} \)
\( = \sqrt{130} \)
In simple words: We first multiply vector \( \vec{a} \) by 2 and vector \( \vec{b} \) by 3, then subtract the results. After getting the new vector, we find its length by taking the square root of the sum of the squares of its parts.

🎯 Exam Tip: When performing operations like scalar multiplication and vector addition/subtraction, ensure you distribute the scalar correctly to all components of the vector. Double-check your arithmetic, especially with negative numbers.

Question 4. If \( \vec{a} \), \( \vec{b} \) and \( \vec{c} \) have components (1, 1) (2, 3) and (-1, 4) respectively, find the components of \( 2 \vec{a} - 3 \vec{b} + 5 \vec{c} \).
Answer:
Given the components, we can write the vectors as:
\( \vec{a} = \hat{i} + \hat{j} \)
\( \vec{b} = 2 \hat{i} + 3 \hat{j} \)
\( \vec{c} = -\hat{i} + 4 \hat{j} \)
Now, we need to calculate \( 2 \vec{a} - 3 \vec{b} + 5 \vec{c} \). This involves multiplying each vector by its scalar and then adding or subtracting the resulting vectors.
\( 2 \vec{a} - 3 \vec{b} + 5 \vec{c} = 2(\hat{i} + \hat{j}) - 3(2 \hat{i} + 3 \hat{j}) + 5(-\hat{i} + 4 \hat{j}) \)
\( = 2 \hat{i} + 2 \hat{j} - 6 \hat{i} - 9 \hat{j} - 5 \hat{i} + 20 \hat{j} \)
Group the \( \hat{i} \) terms and \( \hat{j} \) terms separately.
\( = (2 - 6 - 5)\hat{i} + (2 - 9 + 20)\hat{j} \)
\( = -9 \hat{i} + 13 \hat{j} \)
Thus, the components of \( 2 \vec{a} - 3 \vec{b} + 5 \vec{c} \) are (-9, 13). This means the final vector has an x-component of -9 and a y-component of 13.
In simple words: We are given three vectors as their x and y parts. We multiply each vector by a number and then combine them by adding or subtracting their x-parts together and their y-parts together. The final answer is the new x and y parts of the combined vector.

🎯 Exam Tip: Pay close attention to the signs when distributing scalar multiples and when combining the corresponding components of the vectors. A single sign error can lead to an incorrect final vector.

Question 5. Find a unit vector parallel to \( 3 \hat{i} + 4 \hat{j} \).
Answer:
Let the given vector be \( \vec{a} = 3 \hat{i} + 4 \hat{j} \).
First, we need to find the magnitude of \( \vec{a} \).
\( \implies |\vec{a}| = \sqrt{3^2+4^2} \)
\( = \sqrt{9+16} \)
\( = \sqrt{25} \)
\( = 5 \)
A unit vector in the same direction as \( \vec{a} \) is found by dividing the vector by its magnitude. A unit vector has a length of 1.
\( \implies \) Unit vector parallel to \( \vec{a} \)
\( = \pm \hat{a} = \pm \frac{\vec{a}}{|\vec{a}|} \)
\( = \pm \frac{(3 \hat{i}+4 \hat{j})}{5} \)
In simple words: To find a unit vector (a vector with length 1) that points in the same direction as another vector, first find the length of the original vector. Then, divide each part of the original vector by its length. We use \( \pm \) because a vector can be parallel in two opposite directions.

🎯 Exam Tip: A unit vector always has a magnitude of 1. Remember to include the \( \pm \) sign when asked for a vector *parallel* to a given vector, as it could be in the exact same direction or the opposite direction.

Question 6. Find a unit vector in the direction of \( \hat{i} + \hat{j} \).
Answer:
Let the given vector be \( \vec{a} = \hat{i} + \hat{j} \).
First, we calculate the magnitude of this vector. The components here are 1 and 1.
\( \implies |\vec{a}| = \sqrt{1^2+1^2} \)
\( = \sqrt{1+1} \)
\( = \sqrt{2} \)
A unit vector in the direction of \( \vec{a} \) is found by dividing \( \vec{a} \) by its magnitude. This ensures the new vector has a length of 1 but points in the same way.
\( \implies \) Unit vector in the direction of \( \vec{a} \)
\( = \hat{a} = \frac{\vec{a}}{|\vec{a}|} \)
\( = \frac{\hat{i}+\hat{j}}{\sqrt{2}} \)
In simple words: To get a unit vector that points in the exact same direction, we first find the length of the given vector. Then, we divide the vector by its length. This makes the new vector's length equal to one, but it still points the same way.

🎯 Exam Tip: When asked for a unit vector *in the direction* of a given vector, only the positive result is required, not the \( \pm \) sign, as it specifies a particular direction.

Question 7. Find a vector of magnitude 5 units which is parallel to the vector \( 2 \hat{i} - \hat{j} \).
Answer:
Let the given vector be \( \vec{a} = 2 \hat{i} - \hat{j} \).
First, we find the magnitude of \( \vec{a} \).
\( \implies |\vec{a}| = \sqrt{2^2+(-1)^2} \)
\( = \sqrt{4+1} \)
\( = \sqrt{5} \)
Next, we find the unit vector in the direction of \( \vec{a} \). This vector has a length of 1.
\( \implies \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2 \hat{i} - \hat{j}}{\sqrt{5}} \)
Now, to find a vector of magnitude 5 units parallel to \( \vec{a} \), we multiply the unit vector by the desired magnitude. We include \( \pm \) for parallel directions.
\( \implies \) Required vector \( = \pm 5 \hat{a} \)
\( = \pm 5 \frac{\vec{a}}{|\vec{a}|} \)
\( = \pm 5 \frac{(2 \hat{i}-\hat{j})}{\sqrt{5}} \)
We can simplify this expression by canceling out \( \sqrt{5} \).
\( = \pm \sqrt{5}(2 \hat{i} - \hat{j}) \)
In simple words: We want a vector that has a length of 5 and points in the same line as the given vector. First, we find the length of the given vector. Then, we make it a "unit vector" (length 1) by dividing it by its length. Finally, we multiply this unit vector by 5 to get the required length. Since "parallel" means it can point in either direction, we use plus or minus.

🎯 Exam Tip: The process of creating a vector with a specific magnitude in a particular direction involves two main steps: normalize the given vector to a unit vector, then scale the unit vector by the desired magnitude. Remember the \( \pm \) for parallel vectors.

Question 8. If the components of \( \vec{a} \) are (1, 2) and \( \vec{b} = \hat{i} - \hat{j} + \vec{a} \), what are components of \( 2 \vec{a} - 3 \vec{b} \) ?
Answer:
Given the components, we can write the vector \( \vec{a} \) as:
\( \vec{a} = \hat{i} + 2 \hat{j} \)
Now, substitute \( \vec{a} \) into the expression for \( \vec{b} \):
\( \vec{b} = \hat{i} - \hat{j} + \vec{a} \)
\( = \hat{i} - \hat{j} + (\hat{i} + 2 \hat{j}) \)
Combine the \( \hat{i} \) terms and \( \hat{j} \) terms for \( \vec{b} \).
\( = (1+1)\hat{i} + (-1+2)\hat{j} \)
\( = 2 \hat{i} + \hat{j} \)
Now that we have both \( \vec{a} \) and \( \vec{b} \), we can find the components of \( 2 \vec{a} - 3 \vec{b} \). This involves multiplying each vector by its scalar and then subtracting.
\( 2 \vec{a} - 3 \vec{b} = 2(\hat{i} + 2 \hat{j}) - 3(2 \hat{i} + \hat{j}) \)
\( = 2 \hat{i} + 4 \hat{j} - 6 \hat{i} - 3 \hat{j} \)
Again, combine the \( \hat{i} \) terms and \( \hat{j} \) terms.
\( = (2 - 6)\hat{i} + (4 - 3)\hat{j} \)
\( = -4 \hat{i} + \hat{j} \)
Therefore, the components of \( 2 \vec{a} - 3 \vec{b} \) are (-4, 1). This shows the x and y values for the final combined vector.
In simple words: First, we write vector \( \vec{a} \) from its parts. Then, we use \( \vec{a} \) to find vector \( \vec{b} \). Once we have both \( \vec{a} \) and \( \vec{b} \), we multiply them by the given numbers and subtract to find the x and y parts of the final vector.

🎯 Exam Tip: When one vector is defined in terms of another (e.g., \( \vec{b} = \hat{i} - \hat{j} + \vec{a} \)), always calculate the full expression for that vector first before substituting it into further calculations. This helps avoid errors.

Question 9. Find the terminal point of the vector \( \overrightarrow{P Q} \) whose initial points is P(1, 2) and whose components along x-axis and y-axis are -2 and 3 respectively.
Answer:
Let the initial point P have position vector \( \text{P.V of P} = \hat{i} + 2 \hat{j} \).
The components of vector \( \overrightarrow{PQ} \) are given as -2 along the x-axis and 3 along the y-axis, so \( \overrightarrow{PQ} = -2 \hat{i} + 3 \hat{j} \).
Let the terminal point Q have coordinates \( (\alpha, \beta) \). Then its position vector is \( \text{P.V of Q} = \alpha \hat{i} + \beta \hat{j} \).
We know that \( \overrightarrow{PQ} = \text{P.V of Q} - \text{P.V of P} \). We can use this formula to find the coordinates of Q.
\( \implies -2 \hat{i} + 3 \hat{j} = (\alpha \hat{i} + \beta \hat{j}) - (\hat{i} + 2 \hat{j}) \)
Combine terms on the right side:
\( -2 \hat{i} + 3 \hat{j} = (\alpha - 1) \hat{i} + (\beta - 2) \hat{j} \)
Now, we compare the coefficients of \( \hat{i} \) and \( \hat{j} \) on both sides of the equation. The x-components must be equal and the y-components must be equal.
For \( \hat{i} \) coefficients:
\( \alpha - 1 = -2 \)
\( \implies \alpha = -2 + 1 \)
\( \implies \alpha = -1 \)
For \( \hat{j} \) coefficients:
\( \beta - 2 = 3 \)
\( \implies \beta = 3 + 2 \)
\( \implies \beta = 5 \)
Thus, the required terminal point Q is (-1, 5). This means Q is located at x-coordinate -1 and y-coordinate 5.
In simple words: We are given the start point of a vector and its direction (how much it moves in x and y). We write down the start point and the vector. Then, we use a formula that says the vector is the end point minus the start point. By comparing the x-parts and y-parts on both sides, we find the exact location of the end point.

🎯 Exam Tip: When comparing vector components, remember that the coefficients of \( \hat{i} \) on both sides of the equation must be equal, and similarly for the coefficients of \( \hat{j} \). Be careful with algebra when solving for unknown coordinates.

Question 10. If the coordinates of the points A and B in a plane are (1, 1) and (1, 2) respectively, find the coordinates of C (in the plane) such that \( \overrightarrow{A B} \) and \( \overrightarrow{B C} \) are equal.
Answer:
Given the coordinates, we write the position vectors:
P.V of A \( = \hat{i} + \hat{j} \)
P.V of B \( = \hat{i} + 2 \hat{j} \)
Let the coordinates of point C be \( (\alpha, \beta) \). Then its position vector is \( \text{P.V of C} = \alpha \hat{i} + \beta \hat{j} \).
We are given that \( \overrightarrow{AB} = \overrightarrow{BC} \).
Using the formula \( \overrightarrow{XY} = \text{P.V of Y} - \text{P.V of X} \):
\( \text{P.V of B} - \text{P.V of A} = \text{P.V of C} - \text{P.V of B} \)
Substitute the position vectors:
\( (\hat{i} + 2 \hat{j}) - (\hat{i} + \hat{j}) = (\alpha \hat{i} + \beta \hat{j}) - (\hat{i} + 2 \hat{j}) \)
Simplify both sides of the equation:
Left side:
\( \hat{i} + 2 \hat{j} - \hat{i} - \hat{j} = (1-1)\hat{i} + (2-1)\hat{j} = 0 \hat{i} + \hat{j} \)
Right side:
\( \alpha \hat{i} + \beta \hat{j} - \hat{i} - 2 \hat{j} = (\alpha - 1) \hat{i} + (\beta - 2) \hat{j} \)
So, the equation becomes:
\( 0 \hat{i} + \hat{j} = (\alpha - 1) \hat{i} + (\beta - 2) \hat{j} \)
Comparing the coefficients of \( \hat{i} \) and \( \hat{j} \) on both sides:
For \( \hat{i} \) coefficients:
\( \alpha - 1 = 0 \)
\( \implies \alpha = 1 \)
For \( \hat{j} \) coefficients:
\( \beta - 2 = 1 \)
\( \implies \beta = 3 \)
Hence, the required coordinates of point C are (1, 3). This means point C has the same x-coordinate as A and B, but a higher y-coordinate.
In simple words: We know the start and middle points (A and B). We want to find a third point (C) such that the vector from A to B is the same as the vector from B to C. We write the points as vectors and use the formula: vector = end point - start point. By solving the equation, we find the x and y values for point C.

🎯 Exam Tip: This problem demonstrates how vector equality implies equality of their respective components. Carefully set up the vector equations and then equate coefficients of \( \hat{i} \) and \( \hat{j} \) to solve for unknown coordinates.

Question 11. If A(1, 1), B(2, 4), C(-1, 2) and D(-1, 3) are the points in a plane, find a point P in the same plane such that \( \overrightarrow{A P} = \overrightarrow{A B} + \overrightarrow{C D} \).
Answer:
First, we write the position vectors for the given points:
P.V of A \( = \hat{i} + \hat{j} \)
P.V of B \( = 2\hat{i} + 4\hat{j} \)
P.V of C \( = -\hat{i} + 2\hat{j} \)
P.V of D \( = -\hat{i} + 3\hat{j} \)
Let the coordinates of point P be \( (\alpha, \beta) \). Then its position vector is \( \text{P.V of P} = \alpha \hat{i} + \beta \hat{j} \).
The given equation is \( \overrightarrow{AP} = \overrightarrow{AB} + \overrightarrow{CD} \).
We know that \( \overrightarrow{XY} = \text{P.V of Y} - \text{P.V of X} \). Applying this:
\( \text{P.V of P} - \text{P.V of A} = (\text{P.V of B} - \text{P.V of A}) + (\text{P.V of D} - \text{P.V of C}) \)
Now, substitute the position vectors into the equation:
\( (\alpha \hat{i} + \beta \hat{j}) - (\hat{i} + \hat{j}) = (2\hat{i} + 4\hat{j} - (\hat{i} + \hat{j})) + (-\hat{i} + 3\hat{j} - (-\hat{i} + 2\hat{j})) \)
Simplify each part:
Left side: \( (\alpha - 1) \hat{i} + (\beta - 1) \hat{j} \)
Right side, first part: \( (2\hat{i} + 4\hat{j} - \hat{i} - \hat{j}) = (2-1)\hat{i} + (4-1)\hat{j} = \hat{i} + 3\hat{j} \)
Right side, second part: \( (-\hat{i} + 3\hat{j} + \hat{i} - 2\hat{j}) = (-1+1)\hat{i} + (3-2)\hat{j} = 0\hat{i} + \hat{j} \)
So, the equation becomes:
\( (\alpha - 1) \hat{i} + (\beta - 1) \hat{j} = (\hat{i} + 3\hat{j}) + (\hat{j}) \)
\( (\alpha - 1) \hat{i} + (\beta - 1) \hat{j} = \hat{i} + 4\hat{j} \)
Comparing the coefficients of \( \hat{i} \) and \( \hat{j} \) on both sides:
For \( \hat{i} \) coefficients:
\( \alpha - 1 = 1 \)
\( \implies \alpha = 1 + 1 \)
\( \implies \alpha = 2 \)
For \( \hat{j} \) coefficients:
\( \beta - 1 = 4 \)
\( \implies \beta = 4 + 1 \)
\( \implies \beta = 5 \)
Thus, the coordinates of point P are (2, 5). This means point P is located at x-coordinate 2 and y-coordinate 5.
In simple words: We are given four points and an equation about vectors between them. First, we write each point as a position vector. Then, we use the rule that a vector between two points is found by subtracting their position vectors. We plug these into the main equation and solve for the unknown coordinates of point P by comparing the x and y parts on both sides.

🎯 Exam Tip: Break down complex vector equations into smaller, manageable steps. Calculate each vector (\( \overrightarrow{AB} \), \( \overrightarrow{CD} \), etc.) individually first, then substitute them back into the main equation to avoid computational errors.

Question 12. Prove (by vectors) that the points (-3, 5), (-2, 3) and (4, -9) are collinear.
Answer:
Let the given points be A(-3, 5), B(-2, 3) and C(4, -9). Let O be the origin (0, 0).
The position vectors of the points are:
\( \overrightarrow{OA} = -3\hat{i} + 5\hat{j} \)
\( \overrightarrow{OB} = -2\hat{i} + 3\hat{j} \)
\( \overrightarrow{OC} = 4\hat{i} - 9\hat{j} \)
To prove collinearity using vectors, we can show that two vectors formed by these points are parallel and share a common point. We will find vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).
Vector \( \overrightarrow{AB} \):
\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \)
\( = (-2\hat{i} + 3\hat{j}) - (-3\hat{i} + 5\hat{j}) \)
\( = -2\hat{i} + 3\hat{j} + 3\hat{i} - 5\hat{j} \)
\( = \hat{i} - 2\hat{j} \)
Vector \( \overrightarrow{AC} \):
\( \overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} \)
\( = (4\hat{i} - 9\hat{j}) - (-3\hat{i} + 5\hat{j}) \)
\( = 4\hat{i} - 9\hat{j} + 3\hat{i} - 5\hat{j} \)
\( = 7\hat{i} - 14\hat{j} \)
Now, we check if \( \overrightarrow{AC} \) is a scalar multiple of \( \overrightarrow{AB} \):
\( \overrightarrow{AC} = 7\hat{i} - 14\hat{j} \)
\( = 7(\hat{i} - 2\hat{j}) \)
We can see that \( \overrightarrow{AC} = 7 \overrightarrow{AB} \).
Since \( \overrightarrow{AC} \) is a scalar multiple of \( \overrightarrow{AB} \), the vectors \( \overrightarrow{AC} \) and \( \overrightarrow{AB} \) are parallel. Also, both vectors share a common point A.
Therefore, points A, B, and C are collinear. This means they all lie on the same straight line.
In simple words: To show three points are in a straight line, we find two vectors using these points (like from A to B, and A to C). If one vector is just a scaled version of the other (e.g., one is twice as long as the other), it means they point in the same direction. Since they also share a common starting point (A), all three points must be on the same line.

🎯 Exam Tip: To prove collinearity using vectors, a common method is to show that \( \overrightarrow{AB} = k \overrightarrow{BC} \) (or \( \overrightarrow{AC} = k \overrightarrow{AB} \)) for some scalar \( k \), and that they share a common point. This indicates they lie on the same line.

Question 13. If \( \vec{a} = 7 \hat{i} - 6 \hat{j} \) and \( \vec{b} = 3 \hat{i} + 4 \hat{j} \), find \( \vec{a} + \vec{b} \) and determine a unit vector in the direction of the vector \( \vec{a} + \vec{b} \).
Answer:
Given \( \vec{a} = 7 \hat{i} - 6 \hat{j} \) and \( \vec{b} = 3 \hat{i} + 4 \hat{j} \).
First, we find the sum of the two vectors, \( \vec{a} + \vec{b} \). This involves adding their corresponding components.
\( \implies \vec{a} + \vec{b} = (7 \hat{i} - 6 \hat{j}) + (3 \hat{i} + 4 \hat{j}) \)
\( = (7+3)\hat{i} + (-6+4)\hat{j} \)
\( = 10 \hat{i} - 2 \hat{j} \)
Next, we need to find a unit vector in the direction of \( (\vec{a} + \vec{b}) \). To do this, we first calculate the magnitude of the resultant vector.
\( \implies |\vec{a} + \vec{b}| = \sqrt{10^2+(-2)^2} \)
\( = \sqrt{100+4} \)
\( = \sqrt{104} \)
We can simplify \( \sqrt{104} \) as \( \sqrt{4 \times 26} = 2 \sqrt{26} \).
Finally, the unit vector in the direction of \( \vec{a} + \vec{b} \) is the vector itself divided by its magnitude.
\( \implies \text{Required unit vector} = \frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|} \)
\( = \frac{10 \hat{i}-2 \hat{j}}{2 \sqrt{26}} \)
We can divide both terms in the numerator by 2.
\( = \frac{5 \hat{i}- \hat{j}}{\sqrt{26}} \)
\( = \frac{5}{\sqrt{26}} \hat{i} - \frac{1}{\sqrt{26}} \hat{j} \)
In simple words: First, we add the two given vectors together by combining their x-parts and y-parts. Then, to find a "unit vector" (a vector with length 1) pointing in the same direction as this new combined vector, we find the length of the new vector and divide the new vector by its length.

🎯 Exam Tip: Always simplify square roots in magnitudes if possible (e.g., \( \sqrt{104} = 2\sqrt{26} \)). When expressing the unit vector, ensure both components are divided by the magnitude and presented clearly.

Question 14. Prove that the vectors \( \vec{a} = -4 \hat{i} - \hat{j} \), \( \vec{b} = \hat{i} - 4 \hat{j} \) and \( \vec{c} = 3 \hat{i} + 5 \hat{j} \) form a right angled-triangle.
Answer:
For three vectors to form the sides of a triangle, their sum must be the zero vector (meaning they form a closed loop).
First, let's check if \( \vec{a} + \vec{b} + \vec{c} = \overrightarrow{0} \).
\( \vec{a} + \vec{b} + \vec{c} = (-4 \hat{i} - \hat{j}) + (\hat{i} - 4 \hat{j}) + (3 \hat{i} + 5 \hat{j}) \)
\( = (-4+1+3)\hat{i} + (-1-4+5)\hat{j} \)
\( = (0)\hat{i} + (0)\hat{j} \)
\( = \overrightarrow{0} \)
Since the sum is the zero vector, these three vectors can form the sides of a triangle. Now, we need to determine if it's a right-angled triangle by checking if the Pythagorean theorem applies to their magnitudes.
Calculate the magnitude of each vector:
Magnitude of \( \vec{a} \): \( |\vec{a}| = \sqrt{(-4)^2+(-1)^2} = \sqrt{16+1} = \sqrt{17} \)
Magnitude of \( \vec{b} \): \( |\vec{b}| = \sqrt{1^2+(-4)^2} = \sqrt{1+16} = \sqrt{17} \)
Magnitude of \( \vec{c} \): \( |\vec{c}| = \sqrt{3^2+5^2} = \sqrt{9+25} = \sqrt{34} \)
Now, let's check if the square of the longest side's magnitude equals the sum of the squares of the other two sides' magnitudes.
\( |\vec{a}|^2 = (\sqrt{17})^2 = 17 \)
\( |\vec{b}|^2 = (\sqrt{17})^2 = 17 \)
\( |\vec{c}|^2 = (\sqrt{34})^2 = 34 \)
We observe that \( |\vec{a}|^2 + |\vec{b}|^2 = 17 + 17 = 34 \).
And this is equal to \( |\vec{c}|^2 = 34 \).
Since \( |\vec{a}|^2 + |\vec{b}|^2 = |\vec{c}|^2 \), the vectors satisfy the Pythagorean theorem. Therefore, the triangle formed by these vectors is a right-angled triangle. This indicates that the angle between vectors \( \vec{a} \) and \( \vec{b} \) is 90 degrees.
In simple words: First, we check if the three vectors can form a triangle by adding them up; if the sum is zero, they can. Then, we find the length of each vector. If the square of the longest length is equal to the sum of the squares of the other two lengths, just like in a right-angled triangle (Pythagorean theorem), then the vectors indeed form a right-angled triangle.

🎯 Exam Tip: To prove that vectors form a right-angled triangle, first verify they form a closed loop (\( \vec{a}+\vec{b}+\vec{c} = \overrightarrow{0} \)). Then, calculate the magnitudes of all three vectors and check if the Pythagorean relation \( a^2+b^2=c^2 \) holds for their squared magnitudes. The longest side's square should equal the sum of the squares of the other two.

Question 15. Show that the points \( \vec{a} = -3 \sqrt{3} \hat{i} - 3 \hat{j} \), \( \vec{b} = 6 \hat{j} \) \( \vec{c} = 3 \sqrt{3} \hat{i} - 3 \hat{j} \) form the sides of an isosceles triangle.
Answer:
Let the given vectors be the position vectors of points A, B, and C with respect to the origin O. So, \( \overrightarrow{OA} = -3 \sqrt{3} \hat{i} - 3 \hat{j} \), \( \overrightarrow{OB} = 6 \hat{j} \), and \( \overrightarrow{OC} = 3 \sqrt{3} \hat{i} - 3 \hat{j} \).
To form a triangle, the sum of the vectors representing its sides must be zero. Let's assume these are actual side vectors. If they are position vectors, we need to find the vectors between the points.
Let's define the side vectors of the triangle as \( \overrightarrow{AB} \), \( \overrightarrow{BC} \), and \( \overrightarrow{CA} \).
\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \)
\( = (6 \hat{j}) - (-3 \sqrt{3} \hat{i} - 3 \hat{j}) \)
\( = 6 \hat{j} + 3 \sqrt{3} \hat{i} + 3 \hat{j} \)
\( = 3 \sqrt{3} \hat{i} + 9 \hat{j} \)
Magnitude of \( \overrightarrow{AB} \):
\( |\overrightarrow{AB}| = \sqrt{(3 \sqrt{3})^2 + 9^2} \)
\( = \sqrt{(9 \times 3) + 81} \)
\( = \sqrt{27 + 81} \)
\( = \sqrt{108} \)
\( = \sqrt{36 \times 3} = 6 \sqrt{3} \)

\( \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} \)
\( = (3 \sqrt{3} \hat{i} - 3 \hat{j}) - (6 \hat{j}) \)
\( = 3 \sqrt{3} \hat{i} - 9 \hat{j} \)
Magnitude of \( \overrightarrow{BC} \):
\( |\overrightarrow{BC}| = \sqrt{(3 \sqrt{3})^2 + (-9)^2} \)
\( = \sqrt{(9 \times 3) + 81} \)
\( = \sqrt{27 + 81} \)
\( = \sqrt{108} \)
\( = 6 \sqrt{3} \)

\( \overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} \)
\( = (-3 \sqrt{3} \hat{i} - 3 \hat{j}) - (3 \sqrt{3} \hat{i} - 3 \hat{j}) \)
\( = -3 \sqrt{3} \hat{i} - 3 \hat{j} - 3 \sqrt{3} \hat{i} + 3 \hat{j} \)
\( = -6 \sqrt{3} \hat{i} \)
Magnitude of \( \overrightarrow{CA} \):
\( |\overrightarrow{CA}| = \sqrt{(-6 \sqrt{3})^2 + 0^2} \)
\( = \sqrt{36 \times 3} \)
\( = \sqrt{108} \)
\( = 6 \sqrt{3} \)
We found that \( |\overrightarrow{AB}| = 6 \sqrt{3} \), \( |\overrightarrow{BC}| = 6 \sqrt{3} \), and \( |\overrightarrow{CA}| = 6 \sqrt{3} \).
Since all three sides have equal magnitudes, the triangle formed by these points is an equilateral triangle. An equilateral triangle is a special type of isosceles triangle where all three sides are equal, which means it also satisfies the condition of having at least two sides equal. Therefore, it is an isosceles triangle.
In simple words: We find the lengths of the three sides of the triangle formed by these points. If at least two of these side lengths are the same, then it's an isosceles triangle. In this case, all three sides turned out to be the same length, making it an equilateral triangle, which is also a type of isosceles triangle.

🎯 Exam Tip: To prove a triangle is isosceles using position vectors, calculate the magnitudes of the three vectors formed by pairs of points (e.g., \( |\overrightarrow{AB}| \), \( |\overrightarrow{BC}| \), \( |\overrightarrow{CA}| \)). If any two of these magnitudes are equal, the triangle is isosceles.

Question 16. Show that the three points with position vectors \( 2 \hat{i} + 3 \hat{j} \), \( 3 \hat{i} + \frac{9}{4} \hat{j} \), \( 5 \hat{i} + \frac{3}{4} \hat{j} \) are collinear.
Answer:
Let A, B, and C be the points with the given position vectors:
\( \overrightarrow{OA} = 2\hat{i} + 3\hat{j} \)
\( \overrightarrow{OB} = 3\hat{i} + \frac{9}{4}\hat{j} \)
\( \overrightarrow{OC} = 5\hat{i} + \frac{3}{4}\hat{j} \)
To prove collinearity, we need to show that two vectors formed by these points are parallel and share a common point. We will find \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \).
Vector \( \overrightarrow{AB} \):
\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \)
\( = (3\hat{i} + \frac{9}{4}\hat{j}) - (2\hat{i} + 3\hat{j}) \)
\( = (3-2)\hat{i} + (\frac{9}{4}-3)\hat{j} \)
\( = \hat{i} + (\frac{9-12}{4})\hat{j} \)
\( = \hat{i} - \frac{3}{4}\hat{j} \)
Vector \( \overrightarrow{BC} \):
\( \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} \)
\( = (5\hat{i} + \frac{3}{4}\hat{j}) - (3\hat{i} + \frac{9}{4}\hat{j}) \)
\( = (5-3)\hat{i} + (\frac{3}{4}-\frac{9}{4})\hat{j} \)
\( = 2\hat{i} + (\frac{3-9}{4})\hat{j} \)
\( = 2\hat{i} + (\frac{-6}{4})\hat{j} \)
\( = 2\hat{i} - \frac{3}{2}\hat{j} \)
Now, let's check if \( \overrightarrow{BC} \) is a scalar multiple of \( \overrightarrow{AB} \).
\( \overrightarrow{BC} = 2\hat{i} - \frac{3}{2}\hat{j} \)
We can factor out 2 from the components of \( \overrightarrow{BC} \):
\( \overrightarrow{BC} = 2(\hat{i} - \frac{3}{4}\hat{j}) \)
We observe that \( \overrightarrow{BC} = 2 \overrightarrow{AB} \).
Since \( \overrightarrow{BC} \) is a scalar multiple of \( \overrightarrow{AB} \) (with scalar \( k=2 \)), the vectors \( \overrightarrow{BC} \) and \( \overrightarrow{AB} \) are parallel. Both vectors also share a common point B.
Therefore, points A, B, and C are collinear. This confirms that these three points lie on the same straight line.
In simple words: We are given three points using their position vectors. To show they are in a straight line, we create two vectors between them (like A to B, and B to C). If these two vectors are parallel (meaning one is just a longer or shorter version of the other) and they share a common point, then all three original points must lie on the same straight line.

🎯 Exam Tip: When dealing with fractions in vector components, ensure you find a common denominator correctly before performing addition or subtraction. Always factor out the scalar from one vector to clearly show it's a multiple of the other, proving parallelism.