CBSE Class 12 Mathematics Vectors Algebra MCQs Set 07

Question. The magnitude of the vector \( 6\hat{i} - 2\hat{j} + 3\hat{k} \) is
(a) 1
(b) 5
(c) 7
(d) 12
Answer: (c) 7

Question. If \( \vec{a} + \vec{b} = \hat{i} \) and \( \vec{a} = 2\hat{i} - 2\hat{j} + 2\hat{k} \), then \( |\vec{b}| \) is equal to
(a) \( \sqrt{14} \)
(b) 3
(c) \( \sqrt{12} \)
(d) \( \sqrt{17} \)
Answer: (b) 3

Question. In \( \Delta ABC \), \( \vec{AB} = \hat{i} + \hat{j} + 2\hat{k} \) and \( \vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k} \). If \( D \) is the mid-point of BC, then vector \( \vec{AD} \) is equal to
(a) \( 4\hat{i} + 6\hat{k} \)
(b) \( 2\hat{i} - 2\hat{j} + 2\hat{k} \)
(c) \( \hat{i} - \hat{j} + \hat{k} \)
(d) \( 2\hat{i} + 3\hat{k} \)
Answer: (d) \( 2\hat{i} + 3\hat{k} \)

Question. \( ABCD \) is a rhombus whose diagonals intersect at \( E \). Then, \( \vec{EA} + \vec{EB} + \vec{EC} + \vec{ED} \) is equal to
(a) \( \vec{0} \)
(b) \( \vec{AD} \)
(c) \( 2\vec{BD} \)
(d) \( 2\vec{AD} \)
Answer: (a) \( \vec{0} \)

Question. Two vectors \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \) are collinear, if
(a) \( a_1 b_1 + a_2 b_2 + a_3 b_3 = 0 \)
(b) \( \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \)
(c) \( a_1 = b_1, a_2 = b_2, a_3 = b_3 \)
(d) \( a_1 + a_2 + a_3 = b_1 + b_2 + b_3 \)
Answer: (b) \( \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \)

Question. A unit vector along the vector \( 4\hat{i} - 3\hat{k} \) is
(a) \( \frac{1}{7}(4\hat{i} - 3\hat{k}) \)
(b) \( \frac{1}{5}(4\hat{i} - 3\hat{k}) \)
(c) \( \frac{1}{\sqrt{7}}(4\hat{i} - 3\hat{k}) \)
(d) \( \frac{1}{\sqrt{5}}(4\hat{i} - 3\hat{k}) \)
Answer: (b) \( \frac{1}{5}(4\hat{i} - 3\hat{k}) \)

Question. Unit vector along \( \vec{PQ} \), where coordinates of \( P \) and \( Q \), respectively are \( (2, 1, -1) \) and \( (4, 4, -7) \) is
(a) \( 2\hat{i} + 3\hat{j} - 6\hat{k} \)
(b) \( -2\hat{i} - 3\hat{j} + 6\hat{k} \)
(c) \( \frac{2\hat{i}}{7} - \frac{3\hat{j}}{7} - \frac{6\hat{k}}{7} \)
(d) \( \frac{2\hat{i}}{7} + \frac{3\hat{j}}{7} - \frac{6\hat{k}}{7} \)
Answer: (d) \( \frac{2\hat{i}}{7} + \frac{3\hat{j}}{7} - \frac{6\hat{k}}{7} \)

Question. If a vector makes an angle of \( \frac{\pi}{4} \) with the positive directions of both X-axis and Y-axis, then the angle which it makes with positive Z-axis is
(a) \( \frac{\pi}{4} \)
(b) \( \frac{3\pi}{4} \)
(c) \( \frac{\pi}{2} \)
(d) 0
Answer: (c) \( \frac{\pi}{2} \)

Question. The position vectors of points \( P \) and \( Q \) are \( \vec{p} \) and \( \vec{q} \), respectively. The point \( R \) divides line segment \( PQ \) in the ratio 3 : 1 and \( S \) is the mid-point of line segment \( PR \). The position vector of \( S \) is
(a) \( \frac{\vec{p} + 3\vec{q}}{4} \)
(b) \( \frac{\vec{p} + 3\vec{q}}{8} \)
(c) \( \frac{5\vec{p} + 3\vec{q}}{4} \)
(d) \( \frac{5\vec{p} + 3\vec{q}}{8} \)
Answer: (d) \( \frac{5\vec{p} + 3\vec{q}}{8} \)

Question. Position vector of the mid-point of line segment \( AB \) is \( 3\hat{i} + 2\hat{j} - 3\hat{k} \). If position vector of the point \( A \) is \( 2\hat{i} + 3\hat{j} - 4\hat{k} \), then position vector of the point \( B \) is
(a) \( \frac{5}{2}\hat{i} + \frac{5}{2}\hat{j} - \frac{7}{2}\hat{k} \)
(b) \( 4\hat{i} + \hat{j} - 2\hat{k} \)
(c) \( 5\hat{i} + 5\hat{j} - 7\hat{k} \)
(d) \( \frac{\hat{i}}{2} - \frac{\hat{j}}{2} + \frac{\hat{k}}{2} \)
Answer: (b) \( 4\hat{i} + \hat{j} - 2\hat{k} \)

Question. If two vectors \( \vec{a} \) and \( \vec{b} \) are such that \( |\vec{a}| = 2 \), \( |\vec{b}| = 3 \) and \( \vec{a} \cdot \vec{b} = 4 \), then \( |\vec{a} - 2\vec{b}| \) is equal to
(a) \( \sqrt{2} \)
(b) \( 2\sqrt{6} \)
(c) 24
(d) \( 2\sqrt{2} \)
Answer: (b) \( 2\sqrt{6} \)

Question. The value of \( p \) for which the vectors \( 2\hat{i} + p\hat{j} + \hat{k} \) and \( -4\hat{i} - 6\hat{j} + 26\hat{k} \) are perpendicular to each other is
(a) 3
(b) \( -3 \)
(c) \( -\frac{17}{3} \)
(d) \( \frac{17}{3} \)
Answer: (a) 3

Question. If \( \vec{a} = 4\hat{i} + 6\hat{j} \) and \( \vec{b} = 3\hat{j} + 4\hat{k} \), then the vector form of the component of \( \vec{a} \) along \( \vec{b} \) is
(a) \( \frac{18}{5}(3\hat{i} + 4\hat{k}) \)
(b) \( \frac{18}{25}(3\hat{j} + 4\hat{k}) \)
(c) \( \frac{18}{\sqrt{5}}(3\hat{i} + 4\hat{k}) \)
(d) \( \frac{18}{25}(2\hat{i} + 4\hat{j}) \)
Answer: (b) \( \frac{18}{25}(3\hat{j} + 4\hat{k}) \)

Question. The value of \( \lambda \), for which two vectors \( 2\hat{i} - \hat{j} + 2\hat{k} \) and \( 3\hat{i} + \lambda\hat{j} + \hat{k} \) are perpendicular is
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (d) 8

Question. If \( \vec{a} \cdot \hat{i} = \vec{a} \cdot (\hat{i} + \hat{j}) = \vec{a} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1 \), then \( \vec{a} \) is
(a) \( \hat{k} \)
(b) \( \hat{i} \)
(c) \( \hat{j} \)
(d) \( \hat{i} + \hat{j} + \hat{k} \)
Answer: (b) \( \hat{i} \)

Question. Let \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \). If \( \vec{b} \) is a vector such that \( \vec{a} \cdot \vec{b} = |\vec{b}|^2 \) and \( |\vec{a} - \vec{b}| = \sqrt{7} \), then \( |\vec{b}| \) is equal to
(a) 7
(b) 14
(c) \( \sqrt{7} \)
(d) 21
Answer: (c) \( \sqrt{7} \)

Question. \( \vec{a} \) and \( \vec{b} \) are two non-zero vectors such that the projection of \( \vec{a} \) on \( \vec{b} \) is 0. The angle between \( \vec{a} \) and \( \vec{b} \) is
(a) \( \frac{\pi}{2} \)
(b) \( \pi \)
(c) \( \frac{\pi}{4} \)
(d) 0
Answer: (a) \( \frac{\pi}{2} \)

Question. If \( \theta \) is the angle between two vectors \( \vec{a} \) and \( \vec{b} \), then \( \vec{a} \cdot \vec{b} \ge 0 \) only when
(a) \( 0 < \theta < \frac{\pi}{2} \)
(b) \( 0 \le \theta \le \frac{\pi}{2} \)
(c) \( 0 < \theta < \pi \)
(d) \( 0 \le \theta \le \pi \)
Answer: (b) \( 0 \le \theta \le \frac{\pi}{2} \)

Question. What is the value of \( \frac{\text{projection of } \vec{a} \text{ on } \vec{b}}{\text{projection of } \vec{b} \text{ on } \vec{a}} \) for vectors \( \vec{a} = 2\hat{i} - 3\hat{j} - 6\hat{k} \) and \( \vec{b} = 2\hat{i} - 2\hat{j} + \hat{k} \)?
(a) \( \frac{3}{7} \)
(b) \( \frac{7}{3} \)
(c) \( \frac{4}{3} \)
(d) \( \frac{3}{4} \)
Answer: (b) \( \frac{7}{3} \)

Question. If \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors along three mutually perpendicular directions, then
(a) \( \hat{i} \cdot \hat{j} = 1 \)
(b) \( \hat{i} \times \hat{j} = 1 \)
(c) \( \hat{i} \cdot \hat{k} = 0 \)
(d) \( \hat{i} \times \hat{k} = \vec{0} \)
Answer: (c) \( \hat{i} \cdot \hat{k} = 0 \)

Question. The sine of the angle between the vectors \( \vec{a} = 3\hat{i} + \hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + \hat{j} + 2\hat{k} \) is
(a) \( \frac{\sqrt{5}}{\sqrt{21}} \)
(b) \( \frac{5}{\sqrt{21}} \)
(c) \( \sqrt{\frac{3}{21}} \)
(d) \( \frac{4}{\sqrt{21}} \)
Answer: (a) \( \frac{\sqrt{5}}{\sqrt{21}} \)

Question. If the angle between the vectors \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{4} \) and \( |\vec{a} \times \vec{b}| = 1 \), then \( \vec{a} \cdot \vec{b} \) is equal to
(a) \( -1 \)
(b) 1
(c) \( \frac{1}{\sqrt{2}} \)
(d) \( \sqrt{2} \)
Answer: (b) 1

Question. If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( \vec{a} \cdot \vec{b} > 0 \) and \( |\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}| \), then the angle between \( \vec{a} \) and \( \vec{b} \) is
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{2\pi}{3} \)
(d) \( \frac{3\pi}{4} \)
Answer: (a) \( \frac{\pi}{4} \)

Question. The value of \( (\hat{i} \times \hat{j}) \cdot \hat{j} + (\hat{j} \times \hat{i}) \cdot \hat{k} \) is
(a) 2
(b) 0
(c) 1
(d) \( -1 \)
Answer: (d) \( -1 \)

Question. For any vector \( \vec{a} \), the value of \( (\vec{a} \times \hat{i})^2 + (\vec{a} \times \hat{j})^2 + (\vec{a} \times \hat{k})^2 \) is
(a) \( \vec{a}^2 \)
(b) \( 3\vec{a}^2 \)
(c) \( 4\vec{a}^2 \)
(d) \( 2\vec{a}^2 \)
Answer: (d) \( 2\vec{a}^2 \)

Question. The area of a triangle with vertices A, B and C is given by
(a) \( |\vec{AB} \times \vec{AC}| \)
(b) \( \frac{1}{2}|\vec{AB} \times \vec{AC}| \)
(c) \( \frac{1}{4}|\vec{AC} \times \vec{AB}| \)
(d) \( \frac{1}{8}|\vec{AC} \times \vec{AB}| \)
Answer: (b) \( \frac{1}{2}|\vec{AB} \times \vec{AC}| \)

Assertion-Reason Based Questions

Question. Assertion (A): The sum of two vectors can be zero.
Reason (R): The vector cancels each other when they are equal and opposite.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A): If \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \), then its direction cosines are \( \frac{1}{\sqrt{14}}, \frac{-2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \).
Reason (R): If \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \), then its direction cosines are \( \frac{a_1}{|\vec{a}|}, \frac{a_2}{|\vec{a}|}, \frac{a_3}{|\vec{a}|} \), where \( |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A): If a vector makes \( \alpha, \beta \) and \( \gamma \) with X, Y and Z-axes respectively, then \( \cos 2\alpha + \cos 2\beta + \cos 2\gamma + 1 = 0 \).
Reason (R): Sum of the square of the direction cosines of a vector is equal to 1.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A): The vectors \( \vec{a} = 6\hat{i} + 2\hat{j} - 8\hat{k} \), \( \vec{b} = 10\hat{i} - 2\hat{j} - 6\hat{k} \) and \( \vec{c} = 4\hat{i} - 4\hat{j} + 2\hat{k} \) represent the sides of a right angled triangle.
Reason (R): Three non-zero vectors of which none of two are collinear forms a triangle, if their resultant is zero vector or sum of any two vectors is equal to the third.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A): If \( \vec{a} \) and \( \vec{b} \) are unit vectors such that \( |\vec{a} + \vec{b}| = \sqrt{3} \), then the angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{3} \).
Reason (R): Angle between vectors \( \vec{a} \) and \( \vec{b} \) is given by \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A): If \( |2\vec{a} + \vec{b}| = |2\vec{a} - \vec{b}| \), then \( \vec{a} \) is parallel to \( \vec{b} \).
Reason (R): Two non-zero vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular, if \( \vec{a} \cdot \vec{b} = 0 \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Question. Assertion (A): If \( \vec{a} = 2\hat{i} - 2\hat{j} + 3\hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \), then \( \vec{a} \cdot \vec{b} = 10 \).
Reason (R): If \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \), then \( |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (b) Both A and R are correct; R is not the correct explanation of A.

Question. Assertion (A): The projection of the vector \( \vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k} \) on the vector \( \vec{b} = \hat{i} + 2\hat{j} + \hat{k} \) is \( \frac{5}{3}\sqrt{6} \).
Reason (R): The projection of vector \( \vec{a} \) on vector \( \vec{b} \) is \( \frac{1}{|\vec{a}|}(\vec{a} \cdot \vec{b}) \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (c) A is correct; R is incorrect.

Question. Assertion (A): Unit vector perpendicular to \( \hat{i} - 2\hat{j} + 3\hat{k} \) and \( \hat{i} + 2\hat{j} - \hat{k} \) is \( \frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} - \hat{k}) \).
Reason (R): For two vectors \( \vec{a} \) and \( \vec{b} \), vector \( \vec{a} \times \vec{b} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and unit vector along a vector \( \vec{a} \) is \( \frac{\vec{a}}{|\vec{a}|} \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Question. Assertion (A): Two adjacent sides of a triangle are represented by \( \vec{a} = 3\hat{i} + 4\hat{j} \) and \( \vec{b} = -5\hat{i} + 7\hat{j} \), then the area of the triangle is \( \frac{41}{2} \) sq units.
Reason (R): Area of the triangle whose two adjacent sides are \( \vec{a} \) and \( \vec{b} \) is \( \frac{1}{2}|\vec{a} \times \vec{b}| \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Case Study Based Questions - I

Ram’s house is situated at Shalimar Bagh at point O, for going to Anant’s house he first travels 8 km by bus in the East. Here, at point A, a hospital is situated. From hospital, Ram takes an auto and goes 6 km in the North, here at point B, school is situated. From school, he travels by bus to reach Anant’s house which is at 30° East, 6 km from point B.

Question. What is the vector distance between Ram’s house and school?
Answer: Let \( \vec{OA} = 8\hat{i} \) and \( \vec{AB} = 6\hat{j} \). Then, the vector distance \( \vec{OB} = \vec{OA} + \vec{AB} = 8\hat{i} + 6\hat{j} \).

Question. How much distance Ram travelled to reach school?
Answer: Distance from Ram’s house to school = \( 8 \text{ km} + 6 \text{ km} = 14 \text{ km} \).

Question. What is the vector distance from school to Anant’s house?
Answer: Let a point D be such that \( CD \perp BD \). Then \( BD = 6 \cos 30^\circ \) and \( DC = 6 \sin 30^\circ \). Vector distance from school to Anant's house is \( \vec{BC} = \vec{BD} + \vec{DC} = 6 \cos 30^\circ \hat{i} + 6 \sin 30^\circ \hat{j} = \frac{6\sqrt{3}}{2}\hat{i} + \frac{6}{2}\hat{j} = 3\sqrt{3}\hat{i} + 3\hat{j} \).

Question. What is the vector distance from Ram’s house to Anant’s house?
Answer: Vector distance from Ram's house to Anant's house is \( \vec{OC} = \vec{OB} + \vec{BC} = (8\hat{i} + 6\hat{j}) + (3\sqrt{3}\hat{i} + 3\hat{j}) = (8 + 3\sqrt{3})\hat{i} + 9\hat{j} \).

Question. Find the distance between Ram’s house to Anant’s house.
Answer: Distance between Ram's house to Anant's house = \( OA + AB + BC = 8 + 6 + 6 = 20 \text{ km} \).

Case Study Based Questions - II

On a certain night, Ziad was observing stars using his telescope. Considering the eyepiece of the telescope as the origin, the position vectors of some of the stars are given below:
\( A(2\hat{i} + 3\hat{j} + 4\hat{k}) \), \( B(5\hat{i} + 6\hat{j} + 8\hat{k}) \), \( C(8\hat{i} + 9\hat{j} + 12\hat{k}) \) and \( D(2\hat{i} + 3\hat{j} - 6\hat{k}) \)

Question. Show that the stars with position vectors A, B and C are collinear.
Answer: Given position vectors of some stars are \( A(2\hat{i} + 3\hat{j} + 4\hat{k}) \), \( B(5\hat{i} + 6\hat{j} + 8\hat{k}) \), and \( C(8\hat{i} + 9\hat{j} + 12\hat{k}) \).
Now, \( \vec{AB} = \vec{OB} - \vec{OA} = 3\hat{i} + 3\hat{j} + 4\hat{k} \).
\( \vec{AC} = \vec{OC} - \vec{OA} = 6\hat{i} + 6\hat{j} + 8\hat{k} \).
We know that vectors \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \) are collinear if \( \frac{b_1}{a_1} = \frac{b_2}{a_2} = \frac{b_3}{a_3} \).
Here, \( \frac{6}{3} = \frac{6}{3} = \frac{8}{4} = 2 \). Hence, \( \vec{AC} = 2\vec{AB} \). Therefore, the position vectors A, B and C are collinear.

Question. Express the vector \( \vec{AD} \) in terms of its components and find the distance between the stars A and D. (Note 1 unit = 1 light year)
Answer: The vector \( \vec{AD} \) in component form is \( \vec{AD} = \vec{OD} - \vec{OA} = 0\hat{i} + 0\hat{j} - 10\hat{k} \).
Distance between A and D = \( \sqrt{0^2 + 0^2 + (-10)^2} = \sqrt{100} = 10 \text{ light years} \).

Question. A star at a certain position is inclined at angles of 45° and 60° with the positive direction of the X-axis and Y-axis, respectively. If the star is 2 light years away from O, find its position vector. (Note 1 unit = 1 light year)
Answer: Let \( l, m \) and \( n \) be the direction cosines of the star with respect to X-axis, Y-axis and Z-axis, respectively.
Then, \( l = \cos 45^\circ = \frac{1}{\sqrt{2}} \) and \( m = \cos 60^\circ = \frac{1}{2} \).
We know that \( l^2 + m^2 + n^2 = 1 \).
\( \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + n^2 = 1 \Rightarrow \frac{1}{2} + \frac{1}{4} + n^2 = 1 \Rightarrow n^2 = \frac{1}{4} \Rightarrow n = \pm \frac{1}{2} \).
Now, the position vector of the star is \( \vec{OP} = |\vec{OP}|(l\hat{i} + m\hat{j} + n\hat{k}) = 2\left(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{2}\hat{j} \pm \frac{1}{2}\hat{k}\right) = \sqrt{2}\hat{i} + \hat{j} \pm \hat{k} \) (since \( |\vec{OP}| = 2 \)).

Case Study Based Questions - III

Teams A, B and C went for playing a tug of war game. Teams A, B and C have attached a rope to a metal ring and are trying to pull the ring into their own area. Team A pulls with force \( \vec{F_1} = 6\hat{i} + 0\hat{j} \) kN, Team B pulls with force \( \vec{F_2} = -4\hat{i} + 4\hat{j} \) kN, and Team C pulls with force \( \vec{F_3} = -3\hat{i} - 3\hat{j} \) kN.

Question. What is the magnitude of the force of team A?
Answer: Magnitude of the force of team A is \( |\vec{F_1}| = \sqrt{6^2 + 0^2} = 6 \text{ kN} \).

Question. Which team will win the game?
Answer: Since the magnitude of the force of team A (\( 6\text{ kN} \)) is greater than the other teams (\( |\vec{F_2}| = 4\sqrt{2}\text{ kN} \approx 5.66\text{ kN} \) and \( |\vec{F_3}| = 3\sqrt{2}\text{ kN} \approx 4.24\text{ kN} \)), therefore team A will win the game.

Question. Find the magnitude of the resultant force exerted by the teams.
Answer: Resultant force, \( \vec{F} = \vec{F_1} + \vec{F_2} + \vec{F_3} = (6\hat{i} + 0\hat{j}) + (-4\hat{i} + 4\hat{j}) + (-3\hat{i} - 3\hat{j}) = -\hat{i} + \hat{j} \text{ kN} \).
Magnitude of the resultant force, \( |\vec{F}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \text{ kN} \).

Question. In what direction is the ring getting pulled?
Answer: Let the resultant force \( \vec{F} = -\hat{i} + \hat{j} \) make an angle \( \theta \) with the positive direction of the X-axis. Its direction cosine along the X-axis is:
\( \cos \theta = \frac{-1}{\sqrt{(-1)^2 + 1^2}} = -\frac{1}{\sqrt{2}} \)
\( \theta = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \Rightarrow \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).
Thus, the ring is getting pulled in the direction making an angle of \( \frac{3\pi}{4} \) (or \( 135^\circ \)) with the positive direction of the X-axis.