Mathematics 11 (Punjab Board)
Exercise 14.1 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 14: Vectors in Space.
Question 1: Let \(\mathbf{u} = 3\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}\), \(\mathbf{v} = \mathbf{i} - 5\mathbf{j} - \mathbf{k}\) and \(\mathbf{w} = -4\mathbf{i} - \mathbf{j} + 7\mathbf{k}\). Find: (i) \(\mathbf{u} + 2\mathbf{v} + \mathbf{w}\) (ii) \(\mathbf{v} - 3\mathbf{w}\) (iii) \(|3\mathbf{v} + \mathbf{w}|\)
Vector addition aur scalar multiplication component-wise hoti hai. Magnitude ke liye \(\sqrt{x^2+y^2+z^2}\) use karenge.
(i)
\[\mathbf{u} + 2\mathbf{v} + \mathbf{w} = (3\mathbf{i}+2\mathbf{j}-5\mathbf{k}) + (2\mathbf{i}-10\mathbf{j}-2\mathbf{k}) + (-4\mathbf{i}-\mathbf{j}+7\mathbf{k}) = \mathbf{i} - 9\mathbf{j}\]
(ii)
\[\mathbf{v} - 3\mathbf{w} = (\mathbf{i}-5\mathbf{j}-\mathbf{k}) - (-12\mathbf{i}-3\mathbf{j}+21\mathbf{k}) = 13\mathbf{i} - 2\mathbf{j} - 22\mathbf{k}\]
(iii)
\[3\mathbf{v} + \mathbf{w} = (3\mathbf{i}-15\mathbf{j}-3\mathbf{k}) + (-4\mathbf{i}-\mathbf{j}+7\mathbf{k}) = -\mathbf{i} - 16\mathbf{j} + 4\mathbf{k}\]
\[|3\mathbf{v} + \mathbf{w}| = \sqrt{(-1)^2 + (-16)^2 + 4^2} = \sqrt{273}\]
Answer: (i) \(\mathbf{i} - 9\mathbf{j}\), (ii) \(13\mathbf{i} - 2\mathbf{j} - 22\mathbf{k}\), (iii) \(\sqrt{273}\)
Question 2: Find the magnitude of the vector \(\mathbf{v} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}\) and write its direction cosines.
Magnitude \(\sqrt{x^2+y^2+z^2}\) se nikalenge. Direction cosines ke liye har component ko magnitude se divide karenge.
\[|\mathbf{v}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9+4+36} = \sqrt{49} = 7\]
\[\cos\alpha = \frac{3}{7}, \quad \cos\beta = -\frac{2}{7}, \quad \cos\gamma = \frac{6}{7}\]
Answer: \(|\mathbf{v}| = 7\); direction cosines \(\frac{3}{7}, -\frac{2}{7}, \frac{6}{7}\)
Question 3: Find \(t\) so that \(|2\mathbf{i} + (t-1)\mathbf{j} + t\mathbf{k}| = \sqrt{13}\).
Magnitude ka formula lagayenge, dono taraf square karke quadratic equation banayenge aur solve karenge.
\[\sqrt{2^2 + (t-1)^2 + t^2} = \sqrt{13} \quad \Rightarrow \quad 4 + t^2 - 2t + 1 + t^2 = 13\]
\[2t^2 - 2t - 8 = 0 \quad \Rightarrow \quad t^2 - t - 4 = 0\]
\[t = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}\]
Answer: \(t = \frac{1 + \sqrt{17}}{2}\) or \(t = \frac{1 - \sqrt{17}}{2}\)
Question 4: Find a unit vector in the direction of \(\mathbf{v} = -\mathbf{i} + 4\mathbf{j} - 8\mathbf{k}\).
Unit vector \(= \frac{\mathbf{v}}{|\mathbf{v}|}\). Pehle magnitude nikalenge phir har component ko divide karenge.
\[|\mathbf{v}| = \sqrt{(-1)^2 + 4^2 + (-8)^2} = \sqrt{1+16+64} = \sqrt{81} = 9\]
\[\hat{\mathbf{v}} = \frac{-\mathbf{i} + 4\mathbf{j} - 8\mathbf{k}}{9} = -\frac{1}{9}\mathbf{i} + \frac{4}{9}\mathbf{j} - \frac{8}{9}\mathbf{k}\]
Answer: \(-\frac{1}{9}\mathbf{i} + \frac{4}{9}\mathbf{j} - \frac{8}{9}\mathbf{k}\)
Question 5: If \(\mathbf{u} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k}\), \(\mathbf{v} = -\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\) and \(\mathbf{w} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\), find a unit vector parallel to \(4\mathbf{u} - 3\mathbf{v} + 2\mathbf{w}\).
Pehle \(4\mathbf{u} - 3\mathbf{v} + 2\mathbf{w}\) simplify karke naya vector nikalenge, phir uski magnitude se divide karke unit vector banayenge.
\[4\mathbf{u} - 3\mathbf{v} + 2\mathbf{w} = (8\mathbf{i}+4\mathbf{j}-12\mathbf{k}) + (3\mathbf{i}-12\mathbf{j}-6\mathbf{k}) + (6\mathbf{i}-4\mathbf{j}+2\mathbf{k})\]
\[= 17\mathbf{i} - 12\mathbf{j} - 16\mathbf{k}\]
\[|17\mathbf{i} - 12\mathbf{j} - 16\mathbf{k}| = \sqrt{289 + 144 + 256} = \sqrt{689}\]
\[\text{Unit vector} = \frac{17\mathbf{i} - 12\mathbf{j} - 16\mathbf{k}}{\sqrt{689}}\]
Answer: \(\frac{17\mathbf{i} - 12\mathbf{j} - 16\mathbf{k}}{\sqrt{689}}\)
Question 6: Find a vector whose (i) magnitude is \(5\) and is parallel to \(3\mathbf{i} + 4\mathbf{j} - \mathbf{k}\) (ii) magnitude is \(7\) and is parallel to \(-\mathbf{i} + \mathbf{j} + \mathbf{k}\).
Diye gaye vector ka unit vector nikalenge, phir usay required magnitude se multiply kar denge.
(i)
\[|3\mathbf{i}+4\mathbf{j}-\mathbf{k}| = \sqrt{9+16+1} = \sqrt{26}\]
\[\text{Required vector} = 5 \cdot \frac{3\mathbf{i}+4\mathbf{j}-\mathbf{k}}{\sqrt{26}}\]
(ii)
\[|-\mathbf{i}+\mathbf{j}+\mathbf{k}| = \sqrt{1+1+1} = \sqrt{3}\]
\[\text{Required vector} = 7 \cdot \frac{-\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{3}}\]
Answer: (i) \(\frac{5}{\sqrt{26}}(3\mathbf{i} + 4\mathbf{j} - \mathbf{k})\), (ii) \(\frac{7}{\sqrt{3}}(-\mathbf{i} + \mathbf{j} + \mathbf{k})\)
Question 7: If \(\mathbf{u} = x\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\), \(\mathbf{v} = \mathbf{i} + y\mathbf{j} - 3\mathbf{k}\) and \(\mathbf{w} = -2\mathbf{i} - 3\mathbf{j}\) represent the sides of a triangle, find \(x\) and \(y\).
Triangle ki directed sides ka vector sum zero hota hai: \(\mathbf{u} + \mathbf{v} + \mathbf{w} = \mathbf{0}\). Components compare karke \(x\) aur \(y\) nikalenge.
\[\mathbf{u} + \mathbf{v} + \mathbf{w} = (x+1-2)\mathbf{i} + (2+y-3)\mathbf{j} + (3-3+0)\mathbf{k} = \mathbf{0}\]
\[(x-1)\mathbf{i} + (y-1)\mathbf{j} = \mathbf{0} \quad \Rightarrow \quad x - 1 = 0,\; y - 1 = 0\]
\[\Rightarrow x = 1, \quad y = 1\]
Answer: \(x = 1\), \(y = 1\). Note: the old site key used \(\mathbf{u}+\mathbf{v}=\mathbf{w}\) giving \(x=-3, y=-5\); the standard closed-triangle condition is \(\mathbf{u}+\mathbf{v}+\mathbf{w}=\mathbf{0}\), which gives \(x=1, y=1\).
Question 8: The position vectors of \(A, B, C, D\) are \(\mathbf{u} = \mathbf{i} + 2\mathbf{j} + \mathbf{k}\), \(\mathbf{v} = 7\mathbf{i} + 8\mathbf{j} + 4\mathbf{k}\), \(\mathbf{w} = -\mathbf{i} + \mathbf{k}\), \(\mathbf{z} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\) respectively. Show that \(\overrightarrow{AB}\) is parallel to \(\overrightarrow{CD}\).
Do vectors parallel hote hain agar ek doosre ka scalar multiple ho. \(\overrightarrow{AB} = \mathbf{v} - \mathbf{u}\) aur \(\overrightarrow{CD} = \mathbf{z} - \mathbf{w}\) nikal kar compare karenge.
\[\overrightarrow{AB} = \mathbf{v} - \mathbf{u} = (7-1)\mathbf{i} + (8-2)\mathbf{j} + (4-1)\mathbf{k} = 6\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}\]
\[\overrightarrow{CD} = \mathbf{z} - \mathbf{w} = (1+1)\mathbf{i} + (2-0)\mathbf{j} + (2-1)\mathbf{k} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\]
\[\overrightarrow{AB} = 3(2\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = 3\,\overrightarrow{CD}\]
Answer: \(\overrightarrow{AB} = 3\overrightarrow{CD}\), so they are parallel.
Question 9: Two vectors \(\mathbf{v}, \mathbf{w}\) are parallel if \(\mathbf{v} = c\mathbf{w}\) for some scalar \(c\) (same direction if \(c > 0\), opposite if \(c < 0\)). (a) Find two vectors of length \(2\) parallel to \(\mathbf{v} = 2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}\). (b) Find \(a\) so that \(\mathbf{v} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}\) and \(\mathbf{w} = a\mathbf{i} + 9\mathbf{j} - 12\mathbf{k}\) are parallel. (c) Find a vector of length \(5\) in the direction opposite to \(\mathbf{v} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\). (d) Find \(a, b\) so that \(3\mathbf{i} - \mathbf{j} + 4\mathbf{k}\) and \(a\mathbf{i} + b\mathbf{j} - 2\mathbf{k}\) are parallel.
Unit vector nikal kar required length se multiply karenge (direction ke liye sign adjust karenge). Parallel vectors ke corresponding components ka ratio barabar hota hai.
(a) \(|\mathbf{v}| = \sqrt{4+16+16} = 6\):
\[\pm 2 \cdot \frac{2\mathbf{i}-4\mathbf{j}+4\mathbf{k}}{6} = \pm \frac{2\mathbf{i}-4\mathbf{j}+4\mathbf{k}}{3}\]
(b) Ratios: \(\frac{1}{a} = \frac{-3}{9} = \frac{4}{-12} = -\frac{1}{3}\):
\[\Rightarrow a = -3 \quad (\mathbf{w} = -3\mathbf{v})\]
(c) \(|\mathbf{v}| = \sqrt{14}\); opposite direction means negative sign:
\[-5 \cdot \frac{\mathbf{i}-2\mathbf{j}+3\mathbf{k}}{\sqrt{14}}\]
(d) Ratios: \(\frac{3}{a} = \frac{-1}{b} = \frac{4}{-2} = -2\):
\[\Rightarrow a = -\frac{3}{2}, \quad b = \frac{1}{2}\]
Answer: (a) \(\pm\frac{2\mathbf{i}-4\mathbf{j}+4\mathbf{k}}{3}\); (b) \(a = -3\); (c) \(-\frac{5}{\sqrt{14}}(\mathbf{i}-2\mathbf{j}+3\mathbf{k})\); (d) \(a = -\frac{3}{2}, b = \frac{1}{2}\)
Question 10: A spacecraft moves from \((120, 240, -50)\) to \((130, 210, 80)\) km. Find the magnitude of the displacement vector.
Displacement = final position − initial position. Phir magnitude ka formula lagayenge.
\[\mathbf{d} = (130-120, 210-240, 80+50) = (10, -30, 130)\]
\[|\mathbf{d}| = \sqrt{10^2 + (-30)^2 + 130^2} = \sqrt{100 + 900 + 16900} = \sqrt{17900} = 10\sqrt{179}\]
Answer: \(10\sqrt{179} \approx 133.79\) km
Question 11: Find the direction cosines of: (i) \(\mathbf{u} = -6\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\) (ii) \(\mathbf{v} = 4\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}\) (iii) \(\overrightarrow{PQ}\) where \(P(9,3,13)\) and \(Q(1,6,19)\).
Pehle magnitude nikalenge, phir har component ko magnitude se divide karenge.
(i) \(|\mathbf{u}| = \sqrt{36+9+4} = 7\):
\[-\frac{6}{7}, \quad \frac{3}{7}, \quad \frac{2}{7}\]
(ii) \(|\mathbf{v}| = \sqrt{16+4+25} = 3\sqrt{5}\):
\[\frac{4}{3\sqrt{5}}, \quad \frac{2}{3\sqrt{5}}, \quad -\frac{5}{3\sqrt{5}}\]
(iii) \(\overrightarrow{PQ} = (-8, 3, 6)\), \(|\overrightarrow{PQ}| = \sqrt{64+9+36} = \sqrt{109}\):
\[-\frac{8}{\sqrt{109}}, \quad \frac{3}{\sqrt{109}}, \quad \frac{6}{\sqrt{109}}\]
Answer: (i) \(-\frac{6}{7}, \frac{3}{7}, \frac{2}{7}\); (ii) \(\frac{4}{3\sqrt{5}}, \frac{2}{3\sqrt{5}}, -\frac{5}{3\sqrt{5}}\); (iii) \(-\frac{8}{\sqrt{109}}, \frac{3}{\sqrt{109}}, \frac{6}{\sqrt{109}}\)
Question 12: Which of the following triples can be the direction angles of a single vector? (i) \(45^\circ, 45^\circ, 60^\circ\) (ii) \(30^\circ, 45^\circ, 60^\circ\) (iii) \(45^\circ, 60^\circ, 60^\circ\)
Direction angles ke liye zaroori shart hai \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\). Teenon triples check karenge.
(i)
\[\cos^2 45^\circ + \cos^2 45^\circ + \cos^2 60^\circ = \frac{1}{2} + \frac{1}{2} + \frac{1}{4} = \frac{5}{4} \neq 1\]
(ii)
\[\cos^2 30^\circ + \cos^2 45^\circ + \cos^2 60^\circ = \frac{3}{4} + \frac{1}{2} + \frac{1}{4} = \frac{3}{2} \neq 1\]
(iii)
\[\cos^2 45^\circ + \cos^2 60^\circ + \cos^2 60^\circ = \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1\]
Answer: Only (iii) \(45^\circ, 60^\circ, 60^\circ\) can be direction angles.
