Miscellaneous Exercise, Question 1
Write a unit vector in the XY-plane at \(30^\circ\) to the positive x-axis.
Show solution
- \[\cos 30^\circ\,\hat i + \sin 30^\circ\,\hat j\]
Answer: \(\dfrac{\sqrt3}{2}\hat i + \dfrac12\hat j\)
NCERT Solutions for Class 12 Maths Chapter 10 Miscellaneous Exercise
The Miscellaneous Exercise on Vector Algebra. Mixed practice with components, unit vectors, section formula, dot and cross products; write every vector in \(\hat i, \hat j, \hat k\) form first.
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Miscellaneous Exercise questions and solutions
Miscellaneous Exercise, Question 2
Find the scalar components and magnitude of the vector from P\((x_1, y_1, z_1)\) to Q\((x_2, y_2, z_2)\).
Show solution
- \[\overrightarrow{PQ} = (x_2 - x_1)\hat i + (y_2 - y_1)\hat j + (z_2 - z_1)\hat k\]
Answer: Components \(x_2 - x_1,\ y_2 - y_1,\ z_2 - z_1\); magnitude \[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]
Miscellaneous Exercise, Question 3
A girl walks 4 km west, then 3 km in a direction \(30^\circ\) east of north. Find her displacement from the start.
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- West: \(-4\hat i\). Then \[3(\sin 30^\circ\,\hat i + \cos 30^\circ\,\hat j) = \tfrac32\hat i + \tfrac{3\sqrt3}{2}\hat j\]
- Total: \[-\tfrac52\hat i + \tfrac{3\sqrt3}{2}\hat j\], magnitude \[\sqrt{\tfrac{25}{4} + \tfrac{27}{4}} = \sqrt{13}\]
Answer: \[-\tfrac52\hat i + \tfrac{3\sqrt3}{2}\hat j\] (\(\sqrt{13}\) km)
Miscellaneous Exercise, Question 4
If \(\vec a = \vec b + \vec c\), is \(|\vec a| = |\vec b| + |\vec c|\)? Justify.
Show solution
- Not in general: in the triangle formed by \(\vec b, \vec c, \vec a\), a side is shorter than the sum of the other two, so \(|\vec a| \le |\vec b| + |\vec c|\), with equality only when \(\vec b, \vec c\) point the same way. E.g. \(\hat i + \hat j\) has length \(\sqrt2 < 2\).
Answer: No (only when \(\vec b\) and \(\vec c\) are in the same direction)
Miscellaneous Exercise, Question 5
Find x such that \(x(\hat i + \hat j + \hat k)\) is a unit vector.
Show solution
- \(|x|\sqrt3 = 1\).
Answer: \(x = \pm\dfrac{1}{\sqrt3}\)
Miscellaneous Exercise, Question 6
Find a vector of magnitude 5 parallel to the resultant of \(2\hat i + 3\hat j - \hat k\) and \(\hat i - 2\hat j + \hat k\).
Show solution
- Resultant \(3\hat i + \hat j\), magnitude \(\sqrt{10}\).
Answer: \[\dfrac{5}{\sqrt{10}}(3\hat i + \hat j) = \dfrac{3\sqrt{10}}{2}\hat i + \dfrac{\sqrt{10}}{2}\hat j\]
Miscellaneous Exercise, Question 7
\(\vec a = \hat i + \hat j + \hat k\), \(\vec b = 2\hat i - \hat j + 3\hat k\), \(\vec c = \hat i - 2\hat j + \hat k\). Find a unit vector parallel to \(2\vec a - \vec b + 3\vec c\).
Show solution
- \[2\vec a - \vec b + 3\vec c = 3\hat i - 3\hat j + 2\hat k\], magnitude \(\sqrt{22}\).
Answer: \[\dfrac{1}{\sqrt{22}}(3\hat i - 3\hat j + 2\hat k)\]
Miscellaneous Exercise, Question 8
Show A\((1, -2, -8)\), B\((5, 0, -2)\), C\((11, 3, 7)\) are collinear and find the ratio in which B divides AC.
Show solution
- \[\overrightarrow{AB} = 4\hat i + 2\hat j + 6\hat k\], \[\begin{aligned}\overrightarrow{BC} &= 6\hat i + 3\hat j + 9\hat k \\ &= \tfrac32\overrightarrow{AB}\end{aligned}\]: collinear.
- \(AB : BC = 1 : \tfrac32 = 2 : 3\).
Answer: Collinear; B divides AC internally in the ratio \(2 : 3\)
Miscellaneous Exercise, Question 9
Find R dividing PQ externally in the ratio \(1 : 2\), where P, Q have position vectors \(2\vec a + \vec b\) and \(\vec a - 3\vec b\). Show P is the midpoint of RQ.
Show solution
- \[\begin{aligned}\vec r &= \dfrac{1 \cdot (\vec a - 3\vec b) - 2(2\vec a + \vec b)}{1 - 2} \\ &= 3\vec a + 5\vec b\end{aligned}\]
- Midpoint of RQ: \[\dfrac{(3\vec a + 5\vec b) + (\vec a - 3\vec b)}{2} = 2\vec a + \vec b\], which is P.
Answer: \(\vec r = 3\vec a + 5\vec b\); P is the midpoint of RQ
Miscellaneous Exercise, Question 10
Adjacent sides of a parallelogram are \(2\hat i - 4\hat j + 5\hat k\) and \(\hat i - 2\hat j - 3\hat k\). Find the unit vector along its diagonal and its area.
Show solution
- Diagonal \[\vec a + \vec b = 3\hat i - 6\hat j + 2\hat k\], magnitude 7.
- \[\vec a \times \vec b = 22\hat i + 11\hat j\], magnitude \(\sqrt{605} = 11\sqrt5\).
Answer: Unit vector \(\tfrac17(3\hat i - 6\hat j + 2\hat k)\); area \(11\sqrt5\) square units
Miscellaneous Exercise, Question 11
Show the direction cosines of a vector equally inclined to OX, OY, OZ are \(\pm\left(\tfrac{1}{\sqrt3}, \tfrac{1}{\sqrt3}, \tfrac{1}{\sqrt3}\right)\).
Show solution
- \(l = m = n\) and \(l^2 + m^2 + n^2 = 1\), so \(3l^2 = 1\).
Answer: Shown
Miscellaneous Exercise, Question 12
\(\vec a = \hat i + 4\hat j + 2\hat k\), \(\vec b = 3\hat i - 2\hat j + 7\hat k\), \(\vec c = 2\hat i - \hat j + 4\hat k\). Find \(\vec d\) perpendicular to \(\vec a\) and \(\vec b\) with \(\vec c \cdot \vec d = 15\).
Show solution
- \[\begin{aligned}\vec d &= \lambda(\vec a \times \vec b) \\ &= \lambda(32\hat i - \hat j - 14\hat k)\end{aligned}\]
- \[\begin{aligned}\vec c \cdot \vec d &= \lambda(64 + 1 - 56) \\ &= 9\lambda \\ &= 15\end{aligned}\]
Answer: \[\vec d = \tfrac13(160\hat i - 5\hat j - 70\hat k)\]
Miscellaneous Exercise, Question 13
The dot product of \(\hat i + \hat j + \hat k\) with the unit vector along the sum of \(2\hat i + 4\hat j - 5\hat k\) and \(\lambda\hat i + 2\hat j + 3\hat k\) is 1. Find \(\lambda\).
Show solution
- Sum \[= (2 + \lambda)\hat i + 6\hat j - 2\hat k\]; the condition is \[\dfrac{(2 + \lambda) + 6 - 2}{\sqrt{(2 + \lambda)^2 + 40}} = 1\]
- \[\begin{aligned}&(\lambda + 6)^2 = (\lambda + 2)^2 + 40 \\ \Rightarrow\ &8\lambda = 8\end{aligned}\]
Answer: \(\lambda = 1\)
Miscellaneous Exercise, Question 14
\(\vec a, \vec b, \vec c\) are mutually perpendicular with equal magnitudes. Show that \(\vec a + \vec b + \vec c\) is equally inclined to \(\vec a\), \(\vec b\) and \(\vec c\). (The reprint's wording of this question has a misprint; this is the intended statement.)
Show solution
- \[(\vec a + \vec b + \vec c) \cdot \vec a = |\vec a|^2\] (the other dot products are 0); similarly with \(\vec b\) and \(\vec c\), and \(|\vec a| = |\vec b| = |\vec c|\).
- So \[\cos\alpha = \dfrac{|\vec a|^2}{|\vec a + \vec b + \vec c||\vec a|}\] is the same for all three.
Answer: Shown
Miscellaneous Exercise, Question 15
Prove \((\vec a + \vec b) \cdot (\vec a + \vec b) = |\vec a|^2 + |\vec b|^2\) if and only if \(\vec a, \vec b\) are perpendicular (\(\vec a, \vec b \ne \vec 0\)).
Show solution
- \[(\vec a + \vec b) \cdot (\vec a + \vec b) = |\vec a|^2 + 2\vec a \cdot \vec b + |\vec b|^2\], which equals \(|\vec a|^2 + |\vec b|^2\) exactly when \(\vec a \cdot \vec b = 0\).
Answer: Proved
Miscellaneous Exercise, Question 16
\(\vec a \cdot \vec b \ge 0\) only when: (A) \(0 < \theta < \tfrac{\pi}{2}\) (B) \(0 \le \theta \le \tfrac{\pi}{2}\) (C) \(0 < \theta < \pi\) (D) \(0 \le \theta \le \pi\)
Show solution
- \(\cos\theta \ge 0\) for \[\theta \in \left[0, \tfrac{\pi}{2}\right]\]
Answer: (B)
Miscellaneous Exercise, Question 17
\(\vec a, \vec b\) are unit vectors at angle \(\theta\). \(\vec a + \vec b\) is a unit vector if: (A) \(\tfrac{\pi}{4}\) (B) \(\tfrac{\pi}{3}\) (C) \(\tfrac{\pi}{2}\) (D) \(\tfrac{2\pi}{3}\)
Show solution
- \[\begin{aligned}&|\vec a + \vec b|^2 = 2 + 2\cos\theta = 1 \\ \Rightarrow\ &\cos\theta = -\tfrac12\end{aligned}\]
Answer: (D)
Miscellaneous Exercise, Question 18
\(\hat i \cdot (\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)\) is: (A) 0 (B) \(-1\) (C) 1 (D) 3
Show solution
- \(\hat j \times \hat k = \hat i\), \(\hat i \times \hat k = -\hat j\), \(\hat i \times \hat j = \hat k\): \(1 - 1 + 1\).
Answer: (C) 1
Miscellaneous Exercise, Question 19
\(|\vec a \cdot \vec b| = |\vec a \times \vec b|\) when \(\theta\) is: (A) 0 (B) \(\tfrac{\pi}{4}\) (C) \(\tfrac{\pi}{2}\) (D) \(\pi\)
Show solution
- \(|\cos\theta| = \sin\theta\) in \([0, \pi]\): \(\theta = \tfrac{\pi}{4}\) (or \(\tfrac{3\pi}{4}\)).
Answer: (B)
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