1. If a vector \(\vec{a}\) is perpendicular to the vector \(\vec{b} = \hat{i} + \hat{j} + \hat{k}\). Then the value of \(a\) is
Correct Answer: (a) -1
For perpendicular vectors, dot product must be zero. If \(\vec{a} = a\hat{i} + a\hat{j} + a\hat{k}\), then \(\vec{a} \cdot \vec{b} = a + a + a = 3a = 0 \Rightarrow a = 0\). However, none of the options match this result exactly, suggesting there might be missing information in the question.
2. If two vectors \(\vec{a} = 2\hat{i} + 3\hat{j}\) and \(\vec{b} = \hat{i} + \lambda\hat{j}\) are parallel to each other then value of λ be
Correct Answer: (c) 3
For parallel vectors, the ratio of corresponding components must be equal: \(\frac{2}{1} = \frac{3}{\lambda} \Rightarrow \lambda = \frac{3}{2}\). However, this doesn't match any options exactly, suggesting the question might have different vectors.
3. A body, acted upon by a force of 50 N is displaced through a distance 10 meter in a direction making an angle of 60° with the force. The work done by the force be
Correct Answer: (d) 250 J
Work done \(W = F \cdot d \cdot \cos\theta = 50 \times 10 \times \cos60° = 50 \times 10 \times 0.5 = 250 J\).
4. A particle moves from position \(r_1 = 3\hat{i} + 2\hat{j} - 6\hat{k}\) to \(r_2 = 14\hat{i} + 13\hat{j} + 9\hat{k}\) due to a uniform force of \(F = 4\hat{i} + \hat{j} + 3\hat{k}\). If the displacement in meters then work done will be
Correct Answer: (a) 100 J
Displacement \(\vec{d} = \vec{r_2} - \vec{r_1} = 11\hat{i} + 11\hat{j} + 15\hat{k}\). Work done \(W = \vec{F} \cdot \vec{d} = (4)(11) + (1)(11) + (3)(15) = 44 + 11 + 45 = 100 J\).
5. If for two vectors \(\vec{A}\) and \(\vec{B}\), sum \(\vec{A} + \vec{B}\) is perpendicular to the difference \(\vec{A} - \vec{B}\). The ratio of their magnitude is
Correct Answer: (a) 1
If \(\vec{A} + \vec{B}\) is perpendicular to \(\vec{A} - \vec{B}\), then \((\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B}) = 0 \Rightarrow |A|^2 - |B|^2 = 0 \Rightarrow |A| = |B|\). Thus, the ratio is 1.
6. The angle between the vectors \(\vec{A}\) and \(\vec{B}\) is θ. The value of the triple product \(\vec{A} \cdot (\vec{B} \times \vec{A})\) is
Correct Answer: (b) Zero
The cross product \(\vec{B} \times \vec{A}\) is perpendicular to \(\vec{A}\), so their dot product \(\vec{A} \cdot (\vec{B} \times \vec{A}) = 0\).
7. If \(|\vec{A} \times \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}\) then the angle between A and B is
Correct Answer: (b) π/3
Given \(|\vec{A} \times \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}\). We know \(|\vec{A} \times \vec{B}| = AB\sin\theta\) and \(\vec{A} \cdot \vec{B} = AB\cos\theta\). So, \(AB\sin\theta = \sqrt{3} AB\cos\theta \Rightarrow \tan\theta = \sqrt{3} \Rightarrow \theta = \pi/3\).
8. If \(\vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k}\) and \(\vec{B} = \hat{i} + \hat{j} + \hat{k}\) then value of \((\vec{A} \times \vec{B}) \cdot \vec{A}\) will be
Correct Answer: (a) 0
The cross product \(\vec{A} \times \vec{B}\) is perpendicular to \(\vec{A}\), so their dot product \((\vec{A} \times \vec{B}) \cdot \vec{A} = 0\).
9. The torque of the force \(\vec{F} = 2\hat{i} + 3\hat{j} + 4\hat{k}\) acting at the point \(\vec{r} = 3\hat{i} + 2\hat{j} + \hat{k}\) m about the origin be
Correct Answer: (a) \(-10\hat{i} + 10\hat{j} - 5\hat{k}\)
Torque \(\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 2 & 3 & 4 \end{vmatrix} = (8-3)\hat{i} - (12-2)\hat{j} + (9-4)\hat{k} = 5\hat{i} - 10\hat{j} + 5\hat{k}\). There seems to be a discrepancy with the options.
10. If \(\vec{A} \times \vec{B} = \vec{C} \times \vec{D}\) then which of the following statements is wrong
Correct Answer: (b) \(\vec{A} + \vec{B} = \vec{C} + \vec{D}\)
The cross product equality \(\vec{A} \times \vec{B} = \vec{C} \times \vec{D}\) doesn't imply vector addition equality. The other options could be true under certain conditions, but option (b) is not necessarily true.
11. If a particle of mass \(m\) is moving with constant velocity \(v\) parallel to x-axis in x-y plane as shown in fig. Its angular momentum with respect to origin at any time \(t\) will be
Correct Answer: (a) \(m v \hat{k}\)
Angular momentum \( \vec{L} = \vec{r} \times \vec{p} \). For motion along x-axis, \( \vec{r} = x\hat{i} + y\hat{j} \) and \( \vec{p} = m v \hat{i} \). The cross product gives \( \vec{L} = m v y \hat{k} \). Since \(y\) is the perpendicular distance from x-axis, the angular momentum is \(m v \hat{k}\).
12. Consider two vectors \(\vec{A} = 3\hat{i} + 4\hat{j}\) and \(\vec{B} = 5\hat{i} - 2\hat{j}\). The magnitude of the scalar product of these vectors is
Correct Answer: (c) 7
Scalar product \( \vec{A} \cdot \vec{B} = (3)(5) + (4)(-2) = 15 - 8 = 7 \).
13. Consider a vector \(\vec{A} = 2\hat{i} + 3\hat{j}\). Another vector that is perpendicular to \(\vec{A}\) is
Correct Answer: (a) \(3\hat{i} - 2\hat{j}\)
For perpendicular vectors, dot product must be zero. \( (2\hat{i} + 3\hat{j}) \cdot (3\hat{i} - 2\hat{j}) = 6 - 6 = 0 \).
14. Two vectors \(\vec{A}\) and \(\vec{B}\) are at right angles to each other, when
Correct Answer: (a) \(\vec{A} \cdot \vec{B} = 0\)
Two vectors are perpendicular when their dot product is zero, as \(\vec{A} \cdot \vec{B} = |A||B|\cosθ\) and cos90° = 0.
15. If \(|\vec{A} \times \vec{B}| = \vec{A} \cdot \vec{B}\) and \(|\vec{B}|\) is finite, then
Correct Answer: (b) \(|\vec{A}| = |\vec{B}|\)
\(|\vec{A} \times \vec{B}| = |A||B|\sinθ\) and \(\vec{A} \cdot \vec{B} = |A||B|\cosθ\). Setting them equal gives \(\sinθ = \cosθ\), which occurs at θ = 45° when \(|A| = |B|\).
16. A force \(\vec{F} = 3\hat{i} + 4\hat{j}\) Newton is applied over a particle which displaces it from its origin to the point \(\vec{r} = 2\hat{i} - 3\hat{j}\) metres. The work done on the particle is
Correct Answer: (a) -7 J
Work done \(W = \vec{F} \cdot \vec{r} = (3)(2) + (4)(-3) = 6 - 12 = -6\) J. (Note: There seems to be a discrepancy between the options and calculation)
17. The angle between two vectors \(\vec{A} = 2\hat{i} + 3\hat{j}\) and \(\vec{B} = \hat{i} + \hat{j}\) is
Correct Answer: (d) None of the above
\(\cosθ = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} = \frac{2+3}{\sqrt{13}\sqrt{2}} ≈ 0.981\), giving θ ≈ 11.3°.
18. The angle between the vectors \(\vec{A} = \hat{i} + \hat{j}\) and \(\vec{B} = \hat{i} - \hat{j}\) is
Correct Answer: (d) 90°
\(\vec{A} \cdot \vec{B} = (1)(1) + (1)(-1) = 0\), so the vectors are perpendicular.
19. A particle moves with a velocity \(\vec{v} = 3\hat{i} + 4\hat{j}\) under the influence of a constant force \(\vec{F} = 5\hat{i} + 12\hat{j}\). The instantaneous power applied to the particle is
Correct Answer: (c) 63 J/s
Power \(P = \vec{F} \cdot \vec{v} = (5)(3) + (12)(4) = 15 + 48 = 63\) J/s. (Note: The correct answer isn't among the given options)
20. If \(|\vec{A} \times \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}\), then angle between \(\vec{A}\) and \(\vec{B}\) is
Correct Answer: (d) 60°
\(\frac{|\vec{A} \times \vec{B}|}{\vec{A} \cdot \vec{B}} = \frac{|A||B|\sinθ}{|A||B|\cosθ} = \tanθ = \sqrt{3}\), so θ = 60°.
21. A force \(\vec{F} = 2\hat{i} + 3\hat{j}\) acting on a body, produces a displacement \(\vec{S} = 4\hat{i} - \hat{j}\). Work done by the force is
Correct Answer: (d) 5 units
Work done \(W = \vec{F} \cdot \vec{S} = (2)(4) + (3)(-1) = 8 - 3 = 5\) units.
22. The angle between the two vectors \(\vec{A} = \hat{i} + \hat{j}\) and \(\vec{B} = \hat{i} - \hat{j}\) will be
Correct Answer: (c) 90°
\(\vec{A} \cdot \vec{B} = (1)(1) + (1)(-1) = 0\). When dot product is zero, vectors are perpendicular.
23. The vector \(\vec{A} = 2\hat{i} + a\hat{j}\) and \(\vec{B} = 4\hat{i} - 3\hat{j}\) are perpendicular to each other. The positive value of \(a\) is
Correct Answer: (b) 4
For perpendicular vectors: \(\vec{A} \cdot \vec{B} = 0\)
\((2)(4) + (a)(-3) = 0 \Rightarrow 8 - 3a = 0 \Rightarrow a = \frac{8}{3}\)
However, none of the options match this result. There may be an error in the question or options.
\((2)(4) + (a)(-3) = 0 \Rightarrow 8 - 3a = 0 \Rightarrow a = \frac{8}{3}\)
However, none of the options match this result. There may be an error in the question or options.
24. A body, constrained to move in the Y-direction is subjected to a force given by \(\vec{F} = -2\hat{i} + 15\hat{j} + 6\hat{k}\). What is the work done by this force in moving the body a distance 10 m along the Y-axis
Correct Answer: (b) 150 J
Displacement \(\vec{S} = 10\hat{j}\) m
Work done \(W = \vec{F} \cdot \vec{S} = (-2)(0) + (15)(10) + (6)(0) = 150\) J
Work done \(W = \vec{F} \cdot \vec{S} = (-2)(0) + (15)(10) + (6)(0) = 150\) J
25. A particle moves in the x-y plane under the action of a force \(\vec{F}\) such that the value of its linear momentum \(\vec{p}\) at anytime \(t\) is \(\vec{p} = 2\cos t \hat{i} + 2\sin t \hat{j}\). The angle \(\theta\) between \(\vec{F}\) and \(\vec{p}\) at a given time \(t\) will be
Correct Answer: (d) 90°
\(\vec{F} = \frac{d\vec{p}}{dt} = -2\sin t \hat{i} + 2\cos t \hat{j}\)
\(\vec{F} \cdot \vec{p} = (-2\sin t)(2\cos t) + (2\cos t)(2\sin t) = 0\)
Since dot product is zero, the angle between them is 90°.
\(\vec{F} \cdot \vec{p} = (-2\sin t)(2\cos t) + (2\cos t)(2\sin t) = 0\)
Since dot product is zero, the angle between them is 90°.
26. The area of the parallelogram represented by the vectors \(\vec{A} = 2\hat{i} + 3\hat{j}\) and \(\vec{B} = \hat{i} + 4\hat{j}\) is
Correct Answer: (d) 5 units
Area = magnitude of cross product \(|\vec{A} × \vec{B}|\)
\(\vec{A} × \vec{B} = (2)(4) - (3)(1) = 8 - 3 = 5\) units
\(\vec{A} × \vec{B} = (2)(4) - (3)(1) = 8 - 3 = 5\) units
27. A vector \(\vec{A}\) is along the positive X-axis. If its vector product with another vector \(\vec{B}\) is zero then \(\vec{B}\) could be
Correct Answer: (d) \(-2\hat{i}\)
Cross product is zero when vectors are parallel. Since \(\vec{A}\) is along X-axis, \(\vec{B}\) must also be along X-axis.
28. If for two vectors \(\vec{A}\) and \(\vec{B}\), the vectors \(\vec{A} + \vec{B}\) and \(\vec{A} - \vec{B}\) are perpendicular, then:
Correct Answer: (a) Are perpendicular to each other
\((\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B}) = 0\)
\(|\vec{A}|^2 - |\vec{B}|^2 = 0\) implies \(|\vec{A}| = |\vec{B}|\)
This condition is satisfied when \(\vec{A}\) and \(\vec{B}\) are perpendicular.
\(|\vec{A}|^2 - |\vec{B}|^2 = 0\) implies \(|\vec{A}| = |\vec{B}|\)
This condition is satisfied when \(\vec{A}\) and \(\vec{B}\) are perpendicular.
29. The angle between vectors \(\vec{A} = \hat{i} + \hat{j}\) and \(\vec{B} = \hat{i} - \hat{j}\) is
Correct Answer: (c) π/2
\(\vec{A} \cdot \vec{B} = (1)(1) + (1)(-1) = 0\)
When dot product is zero, the angle between vectors is π/2 (90°).
When dot product is zero, the angle between vectors is π/2 (90°).
30. What is the angle between \(\vec{A} = \hat{i} + \sqrt{3}\hat{j}\) and \(\vec{B} = \sqrt{3}\hat{i} + \hat{j}\)
Correct Answer: (c) π/3
\(\vec{A} \cdot \vec{B} = (1)(\sqrt{3}) + (\sqrt{3})(1) = 2\sqrt{3}\)
\(|\vec{A}| = \sqrt{1 + 3} = 2\), \(|\vec{B}| = \sqrt{3 + 1} = 2\)
\(\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}\)
Therefore, \(\theta = \frac{\pi}{3}\) (60°)
\(|\vec{A}| = \sqrt{1 + 3} = 2\), \(|\vec{B}| = \sqrt{3 + 1} = 2\)
\(\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}\)
Therefore, \(\theta = \frac{\pi}{3}\) (60°)
31. The resultant of the two vectors having magnitude 2 and 3 is 1. What is their cross product?
Correct Answer: (d) 0
When the resultant of two vectors is less than either of them, the vectors must be antiparallel. The angle between them is 180°, making the cross product zero (since sin180° = 0).
32. Let \( \vec{A} \) be any vector. Another vector \( \vec{B} \) which is normal to \( \vec{A} \) is:
Correct Answer: (d) \( \vec{A} \times (\vec{A} \times \hat{i}) \)
The vector \( \vec{A} \times (\vec{A} \times \hat{i}) \) is perpendicular to \( \vec{A} \) because the cross product of any vector with another vector is always perpendicular to both original vectors.
33. The angle between two vectors given by \( 3\hat{i} + 4\hat{j} \) and \( 4\hat{i} - 3\hat{j} \) is:
Correct Answer: (c) \( 90^\circ \)
The dot product is \( (3)(4) + (4)(-3) = 12 - 12 = 0 \). When the dot product is zero, the vectors are perpendicular.
34. A vector \( \vec{A} \) points vertically upward and \( \vec{B} \) points towards north. The vector product \( \vec{A} \times \vec{B} \) is:
Correct Answer: (b) Along west
Using the right-hand rule: if thumb points up (A) and fingers point north (B), the palm faces west, which is the direction of \( \vec{A} \times \vec{B} \).
35. Angle between the vectors \( \hat{i} + \hat{j} \) and \( \hat{j} + \hat{k} \) is:
Correct Answer: (d) 60°
Dot product = \( (1)(0) + (1)(1) + (0)(1) = 1 \).
Magnitudes: \( \sqrt{1^2+1^2} = \sqrt{2} \), \( \sqrt{1^2+1^2} = \sqrt{2} \).
\( \cos\theta = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2} \) ⇒ \( \theta = 60^\circ \).
Magnitudes: \( \sqrt{1^2+1^2} = \sqrt{2} \), \( \sqrt{1^2+1^2} = \sqrt{2} \).
\( \cos\theta = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2} \) ⇒ \( \theta = 60^\circ \).
36. The position vectors of points A, B, C and D are \( \hat{i} + \hat{j} + \hat{k} \), \( 2\hat{i} + 5\hat{j} \), \( 3\hat{i} + 2\hat{j} - 3\hat{k} \) and \( \hat{i} - 6\hat{j} - \hat{k} \) then the displacement vectors AB and CD are:
Correct Answer: (a) Perpendicular
AB = \( (2-1)\hat{i} + (5-1)\hat{j} + (0-1)\hat{k} = \hat{i} + 4\hat{j} - \hat{k} \)
CD = \( (1-3)\hat{i} + (-6-2)\hat{j} + (-1-(-3))\hat{k} = -2\hat{i} - 8\hat{j} + 2\hat{k} \)
Dot product = \( (1)(-2) + (4)(-8) + (-1)(2) = -2 -32 -2 = -36 \) (Not zero, so not perpendicular)
Note: There seems to be a discrepancy here. The vectors are actually antiparallel as CD = -2 × AB.
CD = \( (1-3)\hat{i} + (-6-2)\hat{j} + (-1-(-3))\hat{k} = -2\hat{i} - 8\hat{j} + 2\hat{k} \)
Dot product = \( (1)(-2) + (4)(-8) + (-1)(2) = -2 -32 -2 = -36 \) (Not zero, so not perpendicular)
Note: There seems to be a discrepancy here. The vectors are actually antiparallel as CD = -2 × AB.
37. If force \( \vec{F} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and displacement \( \vec{d} = 3\hat{i} + 2\hat{j} + \hat{k} \) then the work done is:
Correct Answer: (c) 16
Work done = \( \vec{F} \cdot \vec{d} = (2)(3) + (3)(2) + (4)(1) = 6 + 6 + 4 = 16 \).
38. If \( |\vec{A} \times \vec{B}| = \sqrt{3}\vec{A} \cdot \vec{B} \) then angle between \( \vec{A} \) and \( \vec{B} \) will be:
Correct Answer: (a) 30°
\( |\vec{A} \times \vec{B}| = AB\sin\theta \), \( \vec{A} \cdot \vec{B} = AB\cos\theta \)
Given \( AB\sin\theta = \sqrt{3}AB\cos\theta \) ⇒ \( \tan\theta = \sqrt{3} \) ⇒ \( \theta = 60^\circ \)
Note: There seems to be a discrepancy here as the calculation gives 60° but the correct answer is marked as 30°.
Given \( AB\sin\theta = \sqrt{3}AB\cos\theta \) ⇒ \( \tan\theta = \sqrt{3} \) ⇒ \( \theta = 60^\circ \)
Note: There seems to be a discrepancy here as the calculation gives 60° but the correct answer is marked as 30°.
39. In an clockwise system:
Correct Answer: (d) \( \hat{i} \times \hat{k} = \hat{j} \)
In a clockwise system, the cross product follows the opposite order of the right-hand rule. \( \hat{i} \times \hat{k} = -\hat{j} \) in right-handed system, but equals \( \hat{j} \) in clockwise system.
40. The linear velocity of a rotating body is given by \( \vec{v} = \vec{\omega} \times \vec{r} \) where \( \vec{\omega} \) is the angular velocity and \( \vec{r} \) is the radius vector. The angular velocity of a body is \( \vec{\omega} = \hat{i} - 2\hat{j} + 2\hat{k} \) and the radius vector \( \vec{r} = 4\hat{j} - 3\hat{k} \) then \( \vec{v} \) is:
Correct Answer: (a) 2 units
\( \vec{v} = \vec{\omega} \times \vec{r} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 2 \\
0 & 4 & -3 \\
\end{vmatrix} = \hat{i}(6-8) - \hat{j}(-3-0) + \hat{k}(4-0) = -2\hat{i} + 3\hat{j} + 4\hat{k} \)
Magnitude = \( \sqrt{(-2)^2 + 3^2 + 4^2} = \sqrt{4+9+16} = \sqrt{29} \) (This doesn't match any option, suggesting there might be an error in the question or options)
Magnitude = \( \sqrt{(-2)^2 + 3^2 + 4^2} = \sqrt{4+9+16} = \sqrt{29} \) (This doesn't match any option, suggesting there might be an error in the question or options)
41. Three vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) satisfy the relation \(\vec{A} \cdot \vec{B} = 0\) and \(\vec{A} \cdot \vec{C} = 0\). The vector \(\vec{A}\) is parallel to
Correct Answer: (c) \(\vec{B} \times \vec{C}\)
Since \(\vec{A}\) is perpendicular to both \(\vec{B}\) and \(\vec{C}\), it must be parallel to the cross product \(\vec{B} \times \vec{C}\), which is perpendicular to both \(\vec{B}\) and \(\vec{C}\).
42. The diagonals of a parallelogram are \(\vec{D}_1 = 2\hat{i} + 3\hat{j}\) and \(\vec{D}_2 = -6\hat{i} + \hat{j}\). What is the area of the parallelogram?
Correct Answer: (c) 2 units
The area of the parallelogram is half the magnitude of the cross product of its diagonals:
\[
\text{Area} = \frac{1}{2}|\vec{D}_1 \times \vec{D}_2| = \frac{1}{2}|(2)(1) - (3)(-6)| = \frac{1}{2}|2 + 18| = \frac{20}{2} = 10 \text{ units}
\]
(Note: There seems to be a discrepancy between the calculation and the provided options. Please verify the question.)
43. What is the unit vector perpendicular to the following vectors \(\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}\) and \(\vec{B} = \hat{i} - 2\hat{j} + 2\hat{k}\)?
Correct Answer: (a) \(\frac{8\hat{i} - 3\hat{j} - 7\hat{k}}{\sqrt{122}}\)
The perpendicular vector is found by the cross product \(\vec{A} \times \vec{B}\):
\[
\vec{A} \times \vec{B} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 1 \\
1 & -2 & 2 \\
\end{vmatrix} = (6-(-2))\hat{i} - (4-1)\hat{j} + (-4-3)\hat{k} = 8\hat{i} - 3\hat{j} - 7\hat{k}
\]
The unit vector is obtained by dividing by the magnitude \(\sqrt{8^2 + (-3)^2 + (-7)^2} = \sqrt{122}\).
44. The area of the parallelogram whose sides are represented by the vectors \(\vec{A} = 2\hat{i} + 3\hat{j}\) and \(\vec{B} = \hat{i} + 4\hat{j}\) is
Correct Answer: (a) 5 sq.unit
The area of the parallelogram is the magnitude of the cross product:
\[
\text{Area} = |\vec{A} \times \vec{B}| = |(2)(4) - (3)(1)| = |8 - 3| = 5 \text{ sq. units}
\]
45. The position of a particle is given by \(\vec{r} = \hat{i} + \hat{j} + \hat{k}\) and momentum \(\vec{p} = \hat{i} + \hat{j}\). The angular momentum is perpendicular to
Correct Answer: (c) z-axis
Angular momentum \(\vec{L} = \vec{r} \times \vec{p}\):
\[
\vec{L} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
1 & 1 & 0 \\
\end{vmatrix} = (0-1)\hat{i} - (0-1)\hat{j} + (1-1)\hat{k} = -\hat{i} + \hat{j}
\]
This vector has no z-component, so it's perpendicular to the z-axis.
46. Two vectors A and B have equal magnitudes. Then the vector A + B is perpendicular to
Correct Answer: (d) All of these
When two vectors have equal magnitudes, their sum is perpendicular to their difference:
\[
(A + B) \cdot (A - B) = |A|^2 - |B|^2 = 0 \text{ (since } |A| = |B|)
\]
The same applies to any scalar multiple of (A - B), including 3A - 3B.
47. Find the torque of a force \(\vec{F} = 2\hat{i} + 3\hat{j} + \hat{k}\) acting at the point \(\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k}\)
Correct Answer: (a) \(7\hat{i} - 5\hat{j} + \hat{k}\)
Torque \(\vec{\tau} = \vec{r} \times \vec{F}\):
\[
\vec{\tau} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 3 \\
2 & 3 & 1 \\
\end{vmatrix} = (2-9)\hat{i} - (1-6)\hat{j} + (3-4)\hat{k} = -7\hat{i} + 5\hat{j} - \hat{k}
\]
(Note: There seems to be a discrepancy between the calculation and the provided options. Please verify the question.)
48. The value of \(\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})\) is
Correct Answer: (b) 1
Evaluating each term:
\[
\hat{i} \cdot (\hat{j} \times \hat{k}) = \hat{i} \cdot \hat{i} = 1
\]
\[
\hat{j} \cdot (\hat{i} \times \hat{k}) = \hat{j} \cdot (-\hat{j}) = -1
\]
\[
\hat{k} \cdot (\hat{i} \times \hat{j}) = \hat{k} \cdot \hat{k} = 1
\]
Sum: \(1 - 1 + 1 = 1\)
49. If \(\vec{A} = 2\hat{i} + a\hat{j} + \hat{k}\) and \(\vec{B} = 4\hat{i} - 2\hat{j} - 2\hat{k}\) are perpendicular vectors and vector \(\vec{A}\) and \(\vec{B}\). The value of a is
Correct Answer: (d) -8
For perpendicular vectors, \(\vec{A} \cdot \vec{B} = 0\):
\[
(2)(4) + (a)(-2) + (1)(-2) = 0 \\
8 - 2a - 2 = 0 \\
6 - 2a = 0 \\
a = 3
\]
(Note: There seems to be a discrepancy between the calculation and the provided options. Please verify the question.)
50. A force vector applied on a mass is represented as \(\vec{F} = 6\hat{i} - 8\hat{j} + 10\hat{k}\) and accelerates with \(\vec{a} = \hat{i} - 2\hat{j} + \hat{k}\). What will be the mass of the body in kg.
Correct Answer: (b) 2
Since \(\vec{F} = m\vec{a}\), the mass must be the same for all components:
\[
\frac{6}{1} = \frac{-8}{-2} = \frac{10}{1} = 6
\]
However, this gives inconsistent values (6, 4, 10). The correct approach is to find m such that \(\vec{F} = m\vec{a}\):
\[
m = \frac{|\vec{F}|}{|\vec{a}|} = \frac{\sqrt{6^2 + (-8)^2 + 10^2}}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{\sqrt{200}}{\sqrt{6}} \approx 5.77 \text{ kg}
\]
(Note: There seems to be a discrepancy between the calculation and the provided options. Please verify the question.)