Mathematics 11 (Punjab Board)
Exercise 14.2 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 14: Vectors in Space.
Question 1: Find the cosine of the angle \(\theta\) between \(\mathbf{u}\) and \(\mathbf{v}\): (i) \(\mathbf{u} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}\), \(\mathbf{v} = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\) (ii) \(\mathbf{u} = [-3, 2, 5]\), \(\mathbf{v} = [1, 6, -2]\)
Angle ka formula hai \(\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\). Pehle dot product, phir magnitudes nikalenge.
(i)
\[\mathbf{u} \cdot \mathbf{v} = (2)(-1) + (3)(2) + (1)(2) = -2 + 6 + 2 = 6\]
\[|\mathbf{u}| = \sqrt{4+9+1} = \sqrt{14}, \qquad |\mathbf{v}| = \sqrt{1+4+4} = 3\]
\[\cos\theta = \frac{6}{3\sqrt{14}} = \frac{2}{\sqrt{14}}\]
(ii)
\[\mathbf{u} \cdot \mathbf{v} = (-3)(1) + (2)(6) + (5)(-2) = -3 + 12 - 10 = -1\]
\[|\mathbf{u}| = \sqrt{9+4+25} = \sqrt{38}, \qquad |\mathbf{v}| = \sqrt{1+36+4} = \sqrt{41}\]
\[\cos\theta = \frac{-1}{\sqrt{38}\sqrt{41}} = -\frac{1}{\sqrt{1558}}\]
Answer: (i) \(\frac{2}{\sqrt{14}}\), (ii) \(-\frac{1}{\sqrt{1558}}\)
Question 2: If \(\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}\) and \(|\mathbf{a}| = 3, |\mathbf{b}| = 5, |\mathbf{c}| = 7\), find the angle between \(\mathbf{a}\) and \(\mathbf{b}\).
\(\mathbf{a} + \mathbf{b} = -\mathbf{c}\) likh kar dono taraf magnitude ka square lenge. Formula \(|\mathbf{a}+\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2\mathbf{a}\cdot\mathbf{b}\) use hogi.
\[\mathbf{a} + \mathbf{b} = -\mathbf{c} \quad \Rightarrow \quad |\mathbf{a}+\mathbf{b}|^2 = |-\mathbf{c}|^2\]
\[3^2 + 5^2 + 2(3)(5)\cos\theta = 7^2 \quad \Rightarrow \quad 34 + 30\cos\theta = 49\]
\[30\cos\theta = 15 \quad \Rightarrow \quad \cos\theta = \frac{1}{2} \quad \Rightarrow \quad \theta = 60^\circ\]
Answer: \(60^\circ\)
Question 3: If \(|\mathbf{a}| = 3, |\mathbf{b}| = 4\) and \(|\mathbf{a} + \mathbf{b}| = 5\), find the angle between \(\mathbf{a}\) and \(\mathbf{b}\).
Pichle sawal ki tarah \(|\mathbf{a}+\mathbf{b}|^2\) expand karenge aur \(\cos\theta\) solve karenge.
\[|\mathbf{a}+\mathbf{b}|^2 = 25 \quad \Rightarrow \quad 3^2 + 4^2 + 2(3)(4)\cos\theta = 25\]
\[25 + 24\cos\theta = 25 \quad \Rightarrow \quad \cos\theta = 0 \quad \Rightarrow \quad \theta = 90^\circ\]
Answer: \(90^\circ\)
Question 4: Calculate the projection of \(\mathbf{a}\) along \(\mathbf{b}\) and the projection of \(\mathbf{b}\) along \(\mathbf{a}\) when: (i) \(\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), \(\mathbf{b} = \mathbf{i} - 2\mathbf{j} + \mathbf{k}\) (ii) \(\mathbf{a} = 4\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\), \(\mathbf{b} = \mathbf{i} + \mathbf{j} + \mathbf{k}\)
Scalar projection ka formula hai \(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}\) (\(\mathbf{a}\) along \(\mathbf{b}\)). Dot product aur magnitudes nikal kar formula mein rakhenge.
(i)
\[\mathbf{a} \cdot \mathbf{b} = (2)(1) + (3)(-2) + (-1)(1) = -5\]
\[|\mathbf{b}| = \sqrt{6}, \quad |\mathbf{a}| = \sqrt{14}\]
\[\text{proj of } \mathbf{a} \text{ along } \mathbf{b} = \frac{-5}{\sqrt{6}}, \qquad \text{proj of } \mathbf{b} \text{ along } \mathbf{a} = \frac{-5}{\sqrt{14}}\]
(ii)
\[\mathbf{a} \cdot \mathbf{b} = (4)(1) + (-2)(1) + (3)(1) = 5\]
\[|\mathbf{b}| = \sqrt{3}, \quad |\mathbf{a}| = \sqrt{29}\]
\[\text{proj of } \mathbf{a} \text{ along } \mathbf{b} = \frac{5}{\sqrt{3}}, \qquad \text{proj of } \mathbf{b} \text{ along } \mathbf{a} = \frac{5}{\sqrt{29}}\]
Answer: (i) \(-\frac{5}{\sqrt{6}}\) and \(-\frac{5}{\sqrt{14}}\); (ii) \(\frac{5}{\sqrt{3}}\) and \(\frac{5}{\sqrt{29}}\)
Question 5: Find a real number \(\alpha\) so that \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular: (i) \(\mathbf{u} = \alpha\mathbf{i} + 3\mathbf{j} + \mathbf{k}\), \(\mathbf{v} = \mathbf{i} - 2\mathbf{j} + \mathbf{k}\) (ii) \(\mathbf{u} = \alpha\mathbf{i} + 2\alpha\mathbf{j} - \mathbf{k}\), \(\mathbf{v} = \mathbf{i} + \alpha\mathbf{j} + 3\mathbf{k}\)
Perpendicular vectors ka dot product zero hota hai. \(\mathbf{u} \cdot \mathbf{v} = 0\) set karke \(\alpha\) solve karenge.
(i)
\[\mathbf{u} \cdot \mathbf{v} = (\alpha)(1) + (3)(-2) + (1)(1) = \alpha - 5 = 0 \quad \Rightarrow \quad \alpha = 5\]
(ii)
\[\mathbf{u} \cdot \mathbf{v} = (\alpha)(1) + (2\alpha)(\alpha) + (-1)(3) = 2\alpha^2 + \alpha - 3 = 0\]
\[(2\alpha + 3)(\alpha - 1) = 0 \quad \Rightarrow \quad \alpha = 1 \text{ or } \alpha = -\frac{3}{2}\]
Answer: (i) \(\alpha = 5\); (ii) \(\alpha = 1\) or \(\alpha = -\frac{3}{2}\)
Question 6: Find \(z\) so that the triangle with vertices \(A(3, 0, -2)\), \(B(0, 3, 1)\), \(C(1, 1, z)\) is a right triangle with the right angle at \(C\).
Right angle \(C\) par hai, is liye \(\overrightarrow{CA} \cdot \overrightarrow{CB} = 0\). Vectors nikal kar dot product zero set karenge aur quadratic solve karenge.
\[\overrightarrow{CA} = A - C = (2, -1, -2-z), \qquad \overrightarrow{CB} = B - C = (-1, 2, 1-z)\]
\[\overrightarrow{CA} \cdot \overrightarrow{CB} = (2)(-1) + (-1)(2) + (-2-z)(1-z) = 0\]
\[-4 + z^2 + z - 2 = 0 \quad \Rightarrow \quad z^2 + z - 6 = 0\]
\[(z+3)(z-2) = 0 \quad \Rightarrow \quad z = 2 \text{ or } z = -3\]
Answer: \(z = 2\) or \(z = -3\)
Question 7: If \(\mathbf{a}\) and \(\mathbf{b}\) are unit vectors and \(2\theta\) is the angle between them, show that \(\sin\theta = \frac{1}{2}|\mathbf{a} - \mathbf{b}|\).
\(|\mathbf{a} - \mathbf{b}|^2\) expand karenge. Unit vectors hone ki wajah se \(|\mathbf{a}| = |\mathbf{b}| = 1\), aur \(\mathbf{a} \cdot \mathbf{b} = \cos 2\theta\).
\[|\mathbf{a} - \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2\,\mathbf{a}\cdot\mathbf{b} = 1 + 1 - 2\cos 2\theta\]
Using \(\cos 2\theta = 1 - 2\sin^2\theta\):
\[|\mathbf{a} - \mathbf{b}|^2 = 2 - 2(1 - 2\sin^2\theta) = 4\sin^2\theta\]
\[\Rightarrow |\mathbf{a} - \mathbf{b}| = 2\sin\theta \quad \Rightarrow \quad \sin\theta = \frac{1}{2}|\mathbf{a} - \mathbf{b}|\]
Answer: Proved.
Question 8: If \(|\mathbf{a} + \mathbf{b}| = |\mathbf{a} - \mathbf{b}|\), show that \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular.
Dono taraf square lenge. \(|\mathbf{a}|^2\) aur \(|\mathbf{b}|^2\) cancel ho jayenge aur \(\mathbf{a} \cdot \mathbf{b} = 0\) bachega.
\[|\mathbf{a}+\mathbf{b}|^2 = |\mathbf{a}-\mathbf{b}|^2\]
\[|\mathbf{a}|^2 + |\mathbf{b}|^2 + 2\,\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2\,\mathbf{a}\cdot\mathbf{b}\]
\[4\,\mathbf{a}\cdot\mathbf{b} = 0 \quad \Rightarrow \quad \mathbf{a}\cdot\mathbf{b} = 0\]
Since the dot product is zero, \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular.
Answer: Proved.
Question 9: (i) Show that the vectors \(3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\), \(\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}\) and \(2\mathbf{i} + \mathbf{j} - 4\mathbf{k}\) form a right triangle. (ii) Show that the points \(P(4, -1, 2)\), \(Q(1, 3, -1)\), \(R(-2, 4, 6)\) form a right triangle.
(i) mein pehle check karenge ke ek vector baqi do ka sum hai (triangle banta hai), phir dot product se right angle dhoondenge. (ii) mein points se side vectors bana kar teenon dot products check karenge.
(i) Let \(\mathbf{A} = 3\mathbf{i}-2\mathbf{j}+\mathbf{k}\), \(\mathbf{B} = \mathbf{i}-3\mathbf{j}+5\mathbf{k}\), \(\mathbf{C} = 2\mathbf{i}+\mathbf{j}-4\mathbf{k}\):
\[\mathbf{B} + \mathbf{C} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k} = \mathbf{A} \quad \text{(triangle formed)}\]
\[\mathbf{C} \cdot \mathbf{A} = (2)(3) + (1)(-2) + (-4)(1) = 6 - 2 - 4 = 0\]
Since \(\mathbf{C} \perp \mathbf{A}\), the vectors form a right triangle.
(ii)
\[\overrightarrow{PQ} = (-3, 4, -3), \quad \overrightarrow{QR} = (-3, 1, 7), \quad \overrightarrow{RP} = (6, -5, -4)\]
\[\overrightarrow{PQ}\cdot\overrightarrow{QR} = 9+4-21 = -8 \neq 0\]
\[\overrightarrow{QR}\cdot\overrightarrow{RP} = -18-5-28 = -51 \neq 0\]
\[\overrightarrow{RP}\cdot\overrightarrow{PQ} = -18-20+12 = -26 \neq 0\]
No pair of side-vectors is perpendicular, so with the given coordinates the triangle is not right-angled — the question as printed contains a misprint.
Answer: (i) Proved — right angle between \(\mathbf{C}\) and \(\mathbf{A}\). (ii) With the printed coordinates no dot product is zero, so the points do not form a right triangle (misprint in the question).
Question 10: Prove that \(\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta\). Let \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) be unit vectors making angles \(\alpha\) and \(-\beta\) with the positive x-axis.
Do unit vectors ka dot product angle ke cosine ke barabar hota hai. Components se dot product nikal kar identity mil jayegi.
\[\overrightarrow{OA} = \cos\alpha\,\mathbf{i} + \sin\alpha\,\mathbf{j}, \qquad \overrightarrow{OB} = \cos\beta\,\mathbf{i} - \sin\beta\,\mathbf{j}\]
The angle between them is \(\alpha + \beta\), so:
\[\overrightarrow{OA} \cdot \overrightarrow{OB} = (1)(1)\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta\]
Answer: Proved.
Question 11: Prove that in any triangle \(ABC\): (i) \(b = c\cos A + a\cos C\) (ii) \(c = a\cos B + b\cos A\) (iii) \(b^2 = c^2 + a^2 - 2ca\cos B\) (iv) \(c^2 = a^2 + b^2 - 2ab\cos C\). Let vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) lie along \(BC, CA, AB\) respectively.
\(\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}\) se start karenge. Projection formulas ke liye dot product \(\mathbf{b}\) ya \(\mathbf{c}\) ke sath lenge; cosine laws ke liye square karenge.
(i) \(\mathbf{b} = -(\mathbf{a}+\mathbf{c})\); dot with \(\mathbf{b}\):
\[b^2 = -\mathbf{a}\cdot\mathbf{b} - \mathbf{c}\cdot\mathbf{b} = -ab\cos(180^\circ-C) - bc\cos(180^\circ-A) = ab\cos C + bc\cos A\]
\[\Rightarrow b = a\cos C + c\cos A\]
(ii) \(\mathbf{c} = -(\mathbf{a}+\mathbf{b})\); dot with \(\mathbf{c}\):
\[c^2 = ac\cos B + bc\cos A \quad \Rightarrow \quad c = a\cos B + b\cos A\]
(iii) \(\mathbf{b} = -(\mathbf{a}+\mathbf{c})\); square both sides:
\[b^2 = a^2 + c^2 + 2\,\mathbf{a}\cdot\mathbf{c} = a^2 + c^2 + 2ac\cos(180^\circ-B) = c^2 + a^2 - 2ca\cos B\]
(iv) \(\mathbf{c} = -(\mathbf{a}+\mathbf{b})\); square both sides:
\[c^2 = a^2 + b^2 + 2\,\mathbf{a}\cdot\mathbf{b} = a^2 + b^2 + 2ab\cos(180^\circ-C) = a^2 + b^2 - 2ab\cos C\]
Answer: All four results proved.
Question 12: Show that for any vectors \(\mathbf{a}\) and \(\mathbf{b}\): \(|\,|\mathbf{a}| - |\mathbf{b}|\,| \le |\mathbf{a} + \mathbf{b}| \le |\mathbf{a}| + |\mathbf{b}|\).
\(|\mathbf{a}+\mathbf{b}|^2\) expand karenge aur \(\mathbf{a}\cdot\mathbf{b} \le |\mathbf{a}||\mathbf{b}|\) aur \(\mathbf{a}\cdot\mathbf{b} \ge -|\mathbf{a}||\mathbf{b}|\) use karenge. Yehi triangle inequality hai.
\[|\mathbf{a}+\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2\,\mathbf{a}\cdot\mathbf{b}\]
Since \(\mathbf{a}\cdot\mathbf{b} \le |\mathbf{a}||\mathbf{b}|\):
\[|\mathbf{a}+\mathbf{b}|^2 \le (|\mathbf{a}|+|\mathbf{b}|)^2 \quad \Rightarrow \quad |\mathbf{a}+\mathbf{b}| \le |\mathbf{a}|+|\mathbf{b}|\]
Since \(\mathbf{a}\cdot\mathbf{b} \ge -|\mathbf{a}||\mathbf{b}|\):
\[|\mathbf{a}+\mathbf{b}|^2 \ge (|\mathbf{a}|-|\mathbf{b}|)^2 \quad \Rightarrow \quad |\mathbf{a}+\mathbf{b}| \ge |\,|\mathbf{a}|-|\mathbf{b}|\,|\]
Answer: Proved — the triangle inequality for vectors.
Question 13: Find the work done if the constant force \(\mathbf{F} = 2\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}\) moves its point of application from \(P_1(2, -3, 1)\) to \(P_2(7, 5, 3)\).
Work done \(W = \mathbf{F} \cdot \mathbf{d}\), jahan \(\mathbf{d} = P_2 - P_1\) displacement vector hai.
\[\mathbf{d} = P_2 - P_1 = 5\mathbf{i} + 8\mathbf{j} + 2\mathbf{k}\]
\[W = \mathbf{F} \cdot \mathbf{d} = (2)(5) + (5)(8) + (3)(2) = 10 + 40 + 6 = 56\]
Answer: \(56\) units of work
Question 14: A particle acted on by constant forces \(\mathbf{F}_1 = 3\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}\) and \(\mathbf{F}_2 = \mathbf{i} + 4\mathbf{j} - \mathbf{k}\) is displaced from \(A(2, 1, 3)\) to \(B(5, 4, 4)\). Find the work done.
Pehle total force \(\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2\) nikalenge, phir displacement \(\mathbf{d} = B - A\), phir \(W = \mathbf{F} \cdot \mathbf{d}\).
\[\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2 = 4\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}, \qquad \mathbf{d} = B - A = 3\mathbf{i} + 3\mathbf{j} + \mathbf{k}\]
\[W = \mathbf{F} \cdot \mathbf{d} = (4)(3) + (8)(3) + (-4)(1) = 12 + 24 - 4 = 32\]
Answer: \(32\) units of work
Question 15: A particle is displaced from \(A(3, -5, -7)\) to \(B(6, 2, -2)\) under the constant forces \(10\mathbf{i} - \mathbf{j} + 11\mathbf{k}\), \(4\mathbf{i} + 5\mathbf{j} + 9\mathbf{k}\) and \(-2\mathbf{i} + \mathbf{j} - 9\mathbf{k}\). Find the total work done.
Teenon forces ka total nikalenge, phir displacement, phir dot product se work done.
\[\mathbf{F} = (10+4-2)\mathbf{i} + (-1+5+1)\mathbf{j} + (11+9-9)\mathbf{k} = 12\mathbf{i} + 5\mathbf{j} + 11\mathbf{k}\]
\[\mathbf{d} = B - A = 3\mathbf{i} + 7\mathbf{j} + 5\mathbf{k}\]
\[W = \mathbf{F} \cdot \mathbf{d} = (12)(3) + (5)(7) + (11)(5) = 36 + 35 + 55 = 126\]
Answer: \(126\) units of work. Note: some printed keys list \(102\); direct computation gives \(126\).
Question 16: A force of magnitude \(6\) acting parallel to \(4\mathbf{i} + 3\mathbf{j} - \mathbf{k}\) displaces its point of application from \(A(2, -1, 3)\) to \(B(7, 3, 2)\). Find the work done.
Force vector = magnitude × (direction ka unit vector). Phir displacement nikal kar \(W = \mathbf{F} \cdot \mathbf{d}\).
\[\mathbf{F} = 6 \cdot \frac{4\mathbf{i}+3\mathbf{j}-\mathbf{k}}{\sqrt{16+9+1}} = \frac{6}{\sqrt{26}}(4\mathbf{i}+3\mathbf{j}-\mathbf{k})\]
\[\mathbf{d} = B - A = 5\mathbf{i} + 4\mathbf{j} - \mathbf{k}\]
\[W = \mathbf{F} \cdot \mathbf{d} = \frac{6}{\sqrt{26}}\big[(4)(5) + (3)(4) + (-1)(-1)\big] = \frac{6}{\sqrt{26}}(33) = \frac{198}{\sqrt{26}}\]
Answer: \(\frac{198}{\sqrt{26}}\) units of work
