Mathematics 11 (Punjab Board)
Exercise 14.3 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 14: Vectors in Space (Cross Product).
Question 1: Find \(\mathbf{a} \times \mathbf{b}\) and \(\mathbf{b} \times \mathbf{a}\), and show that each is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\): (i) \(\mathbf{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}\), \(\mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}\) (ii) \(\mathbf{a} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\), \(\mathbf{b} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\)
Cross product determinant se nikalenge. Perpendicular hone ki verification dot product zero se hogi. \(\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})\) yaad rakhen.
(i)
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix} = (1-1)\mathbf{i} - (2+1)\mathbf{j} + (-2-1)\mathbf{k} = -3\mathbf{j} - 3\mathbf{k}\]
\[\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b}) = 3\mathbf{j} + 3\mathbf{k}\]
\[\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = (2)(0)+(1)(-3)+(-1)(-3) = 0, \quad \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = (1)(0)+(-1)(-3)+(1)(-3) = 0\]
(ii)
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -3 \\ 3 & -1 & 2 \end{vmatrix} = (4-3)\mathbf{i} - (2+9)\mathbf{j} + (-1-6)\mathbf{k} = \mathbf{i} - 11\mathbf{j} - 7\mathbf{k}\]
\[\mathbf{b} \times \mathbf{a} = -\mathbf{i} + 11\mathbf{j} + 7\mathbf{k}\]
\[\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 1 - 22 + 21 = 0, \quad \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = 3 + 11 - 14 = 0\]
Answer: (i) \(\mathbf{a} \times \mathbf{b} = -3\mathbf{j}-3\mathbf{k}\), \(\mathbf{b} \times \mathbf{a} = 3\mathbf{j}+3\mathbf{k}\); (ii) \(\mathbf{a} \times \mathbf{b} = \mathbf{i}-11\mathbf{j}-7\mathbf{k}\), \(\mathbf{b} \times \mathbf{a} = -\mathbf{i}+11\mathbf{j}+7\mathbf{k}\) — each perpendicular to both original vectors.
Question 2: Find a unit vector perpendicular to the plane of \(\mathbf{a}\) and \(\mathbf{b}\), and find \(\sin\theta\): (i) \(\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), \(\mathbf{b} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\) (ii) \(\mathbf{a} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\), \(\mathbf{b} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}\)
Unit perpendicular vector \(= \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|}\). Angle ke liye \(\sin\theta = \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}\).
(i)
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -1 \\ 1 & -2 & 2 \end{vmatrix} = (6-2)\mathbf{i} - (4+1)\mathbf{j} + (-4-3)\mathbf{k} = 4\mathbf{i} - 5\mathbf{j} - 7\mathbf{k}\]
\[|\mathbf{a} \times \mathbf{b}| = \sqrt{16+25+49} = \sqrt{90} = 3\sqrt{10}\]
\[\text{Unit vector} = \frac{4\mathbf{i}-5\mathbf{j}-7\mathbf{k}}{3\sqrt{10}}, \qquad \sin\theta = \frac{3\sqrt{10}}{\sqrt{14} \cdot 3} = \sqrt{\frac{5}{7}}\]
(ii)
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 2 \\ 2 & 1 & -1 \end{vmatrix} = (1-2)\mathbf{i} - (-3-4)\mathbf{j} + (3+2)\mathbf{k} = -\mathbf{i} + 7\mathbf{j} + 5\mathbf{k}\]
\[|\mathbf{a} \times \mathbf{b}| = \sqrt{1+49+25} = \sqrt{75} = 5\sqrt{3}\]
\[\text{Unit vector} = \frac{-\mathbf{i}+7\mathbf{j}+5\mathbf{k}}{5\sqrt{3}}, \qquad \sin\theta = \frac{5\sqrt{3}}{\sqrt{14}\sqrt{6}} = \frac{5\sqrt{7}}{14}\]
Answer: (i) \(\frac{4\mathbf{i}-5\mathbf{j}-7\mathbf{k}}{3\sqrt{10}}\), \(\sin\theta = \sqrt{5/7}\); (ii) \(\frac{-\mathbf{i}+7\mathbf{j}+5\mathbf{k}}{5\sqrt{3}}\), \(\sin\theta = \frac{5\sqrt{7}}{14}\)
Question 3: Find the area of the triangular region with vertices: (i) \(A(1, 2, 3)\), \(B(-1, 0, 2)\), \(C(2, 1, 0)\) (ii) \(A(0, 1, 2)\), \(B(3, -1, 1)\), \(C(1, 2, -1)\)
Triangle ka area \(= \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|\). Pehle do side vectors nikalenge, phir cross product aur uski magnitude.
(i)
\[\overrightarrow{AB} = -2\mathbf{i}-2\mathbf{j}-\mathbf{k}, \quad \overrightarrow{AC} = \mathbf{i}-\mathbf{j}-3\mathbf{k}\]
\[\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & -2 & -1 \\ 1 & -1 & -3 \end{vmatrix} = 5\mathbf{i} - 7\mathbf{j} + 4\mathbf{k}\]
\[\text{Area} = \frac{1}{2}\sqrt{25+49+16} = \frac{3\sqrt{10}}{2}\]
(ii)
\[\overrightarrow{AB} = 3\mathbf{i}-2\mathbf{j}-\mathbf{k}, \quad \overrightarrow{AC} = \mathbf{i}+\mathbf{j}-3\mathbf{k}\]
\[\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & -1 \\ 1 & 1 & -3 \end{vmatrix} = 7\mathbf{i} + 8\mathbf{j} + 5\mathbf{k}\]
\[\text{Area} = \frac{1}{2}\sqrt{49+64+25} = \frac{\sqrt{138}}{2}\]
Answer: (i) \(\frac{3\sqrt{10}}{2}\) square units; (ii) \(\frac{\sqrt{138}}{2}\) square units
Question 4: Find the area of the parallelogram: (i) with adjacent sides \(\mathbf{u} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{v} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}\) (ii) with vertices \(P(1, 0, 1)\), \(Q(2, 1, 3)\), \(R(-1, 2, 0)\) taking \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\) as adjacent sides.
Parallelogram ka area \(= |\text{(side}_1) \times \text{(side}_2)|\). (ii) mein pehle \(\overrightarrow{PQ}\) aur \(\overrightarrow{PR}\) nikalenge.
(i)
\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 2 \\ 1 & 2 & -1 \end{vmatrix} = (1-4)\mathbf{i} - (-3-2)\mathbf{j} + (6+1)\mathbf{k} = -3\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}\]
\[\text{Area} = \sqrt{9+25+49} = \sqrt{83}\]
(ii)
\[\overrightarrow{PQ} = \mathbf{i}+\mathbf{j}+2\mathbf{k}, \quad \overrightarrow{PR} = -2\mathbf{i}+2\mathbf{j}-\mathbf{k}\]
\[\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2 \\ -2 & 2 & -1 \end{vmatrix} = (-1-4)\mathbf{i} - (-1+4)\mathbf{j} + (2+2)\mathbf{k} = -5\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}\]
\[\text{Area} = \sqrt{25+9+16} = \sqrt{50} = 5\sqrt{2}\]
Answer: (i) \(\sqrt{83}\) square units; (ii) \(5\sqrt{2}\) square units
Question 5: The vectors \(\mathbf{a} = 2\mathbf{i} + \alpha\mathbf{j} - 3\mathbf{k}\) and \(\mathbf{b} = 4\mathbf{i} - 2\mathbf{j} + \beta\mathbf{k}\) are parallel. Find \(\alpha\) and \(\beta\).
Parallel vectors ka cross product zero hota hai. \(\mathbf{a} \times \mathbf{b} = \mathbf{0}\) set karke har component se equations banayenge.
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & \alpha & -3 \\ 4 & -2 & \beta \end{vmatrix} = (\alpha\beta-6)\mathbf{i} - (2\beta+12)\mathbf{j} + (-4-4\alpha)\mathbf{k} = \mathbf{0}\]
\[\alpha\beta - 6 = 0, \quad 2\beta + 12 = 0, \quad -4 - 4\alpha = 0\]
\[\Rightarrow \beta = -6, \quad \alpha = -1 \quad (\text{check: } (-1)(-6) = 6 \checkmark)\]
Answer: \(\alpha = -1\), \(\beta = -6\)
Question 6: Identify whether each pair is parallel, perpendicular, or neither: (i) \(\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\), \(\mathbf{b} = -4\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}\) (ii) \(\mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k}\), \(\mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}\) (iii) \(\mathbf{a} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\), \(\mathbf{b} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}\)
Parallel ke liye cross product zero, perpendicular ke liye dot product zero check karenge.
(i) \(\mathbf{b} = -2\mathbf{a}\):
\[\mathbf{a} \times \mathbf{b} = \mathbf{0} \quad \Rightarrow \quad \text{parallel (opposite directions)}\]
(ii)
\[\mathbf{a} \cdot \mathbf{b} = (3)(1) + (1)(-1) + (-2)(1) = 3 - 1 - 2 = 0 \quad \Rightarrow \quad \text{perpendicular}\]
(iii)
\[\mathbf{a} \cdot \mathbf{b} = 2 - 2 - 3 = -3 \neq 0, \qquad \mathbf{a} \times \mathbf{b} = -\mathbf{i} - 7\mathbf{j} - 5\mathbf{k} \neq \mathbf{0}\]
Neither dot nor cross product is zero.
Answer: (i) parallel; (ii) perpendicular; (iii) neither
Question 7: Prove the following properties of the cross product: (i) \(\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})\) (ii) \(\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}\) (iii) \((k\mathbf{a}) \times \mathbf{b} = k(\mathbf{a} \times \mathbf{b})\)
Determinant ki properties use hongi: rows swap karne se sign badalta hai, aur ek row ka scalar multiple bahar nikala ja sakta hai.
(i) \(\mathbf{b} \times \mathbf{a}\) ke determinant mein do rows swap hain, is liye sign flip hota hai:
\[\mathbf{b} \times \mathbf{a} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \end{vmatrix} = -\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = -(\mathbf{a} \times \mathbf{b})\]
(ii) Determinant rows mein linear hota hai:
\[\mathbf{a} \times (\mathbf{b}+\mathbf{c}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1+c_1 & b_2+c_2 & b_3+c_3 \end{vmatrix} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} + \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\]
(iii) Scalar \(k\) ko determinant ki row se bahar nikal sakte hain:
\[(k\mathbf{a}) \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ ka_1 & ka_2 & ka_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = k\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = k(\mathbf{a} \times \mathbf{b})\]
Answer: All three properties proved.
Question 8: If \(\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}\), show that \(\mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{c} = \mathbf{c} \times \mathbf{a}\).
\(\mathbf{c} = -(\mathbf{a}+\mathbf{b})\) substitute karke cross product ki distributive property use karenge. \(\mathbf{x} \times \mathbf{x} = \mathbf{0}\) yaad rakhen.
\[\mathbf{b} \times \mathbf{c} = \mathbf{b} \times (-(\mathbf{a}+\mathbf{b})) = -\mathbf{b}\times\mathbf{a} - \mathbf{b}\times\mathbf{b} = \mathbf{a}\times\mathbf{b} - \mathbf{0} = \mathbf{a}\times\mathbf{b}\]
\[\mathbf{c} \times \mathbf{a} = (-(\mathbf{a}+\mathbf{b})) \times \mathbf{a} = -\mathbf{a}\times\mathbf{a} - \mathbf{b}\times\mathbf{a} = \mathbf{0} + \mathbf{a}\times\mathbf{b} = \mathbf{a}\times\mathbf{b}\]
Answer: \(\mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{c} = \mathbf{c} \times \mathbf{a}\). Proved.
Question 9: Prove the vector triple product identity \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}\).
Components mein likh kar verify karenge. Pehle \(\mathbf{b} \times \mathbf{c}\) nikalenge, phir \(\mathbf{a}\) ke sath cross product. \(a_1b_1c_1\) add/subtract karne se dot products ban jayenge.
Let \(\mathbf{a} = (a_1,a_2,a_3)\), \(\mathbf{b} = (b_1,b_2,b_3)\), \(\mathbf{c} = (c_1,c_2,c_3)\). Then \(\mathbf{b} \times \mathbf{c} = (b_2c_3-b_3c_2,\; b_3c_1-b_1c_3,\; b_1c_2-b_2c_1)\). The first component of \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\) is:
\[a_2(b_1c_2-b_2c_1) - a_3(b_3c_1-b_1c_3) = a_2b_1c_2 - a_2b_2c_1 - a_3b_3c_1 + a_3b_1c_3\]
Adding and subtracting \(a_1b_1c_1\):
\[= b_1(a_1c_1+a_2c_2+a_3c_3) - c_1(a_1b_1+a_2b_2+a_3b_3) = b_1(\mathbf{a}\cdot\mathbf{c}) - c_1(\mathbf{a}\cdot\mathbf{b})\]
This is the first component of \((\mathbf{a}\cdot\mathbf{c})\mathbf{b} - (\mathbf{a}\cdot\mathbf{b})\mathbf{c}\); the other components follow by cyclic symmetry.
Answer: \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}\). Proved.
Question 10: Prove that \(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta\) using the cross product of unit vectors.
xy-plane mein unit vectors \(\overrightarrow{OA}\) (angle \(\alpha\)) aur \(\overrightarrow{OB}\) (angle \(-\beta\)) lein ge. Unka cross product magnitude \(\sin(\alpha+\beta)\) dega.
\[\overrightarrow{OA} = \cos\alpha\,\mathbf{i} + \sin\alpha\,\mathbf{j}, \qquad \overrightarrow{OB} = \cos\beta\,\mathbf{i} - \sin\beta\,\mathbf{j}\]
The (signed) angle from \(\overrightarrow{OA}\) to \(\overrightarrow{OB}\) is \(-(\alpha+\beta)\), so \(\overrightarrow{OA} \times \overrightarrow{OB} = -\sin(\alpha+\beta)\,\mathbf{k}\). By components:
\[\overrightarrow{OA} \times \overrightarrow{OB} = \big[(\cos\alpha)(-\sin\beta) - (\sin\alpha)(\cos\beta)\big]\mathbf{k} = -\sin(\alpha+\beta)\,\mathbf{k}\]
\[\therefore \sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta\]
Answer: Proved.
Question 11: Establish the relation between the dot and cross products: \(|\mathbf{a} \times \mathbf{b}|^2 + (\mathbf{a} \cdot \mathbf{b})^2 = |\mathbf{a}|^2|\mathbf{b}|^2\).
Dono taraf angle \(\theta\) ki form mein likhenge: \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta\) aur \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta\). Phir \(\sin^2\theta + \cos^2\theta = 1\).
\[|\mathbf{a} \times \mathbf{b}|^2 + (\mathbf{a} \cdot \mathbf{b})^2 = |\mathbf{a}|^2|\mathbf{b}|^2\sin^2\theta + |\mathbf{a}|^2|\mathbf{b}|^2\cos^2\theta\]
\[= |\mathbf{a}|^2|\mathbf{b}|^2(\sin^2\theta + \cos^2\theta) = |\mathbf{a}|^2|\mathbf{b}|^2\]
Answer: \(|\mathbf{a} \times \mathbf{b}|^2 + (\mathbf{a} \cdot \mathbf{b})^2 = |\mathbf{a}|^2|\mathbf{b}|^2\). Proved.
Question 12: Show that \((\mathbf{a} + \mathbf{b}) \times (\mathbf{a} - \mathbf{b}) = 2(\mathbf{b} \times \mathbf{a})\).
Cross product distribute karenge. \(\mathbf{a} \times \mathbf{a} = \mathbf{0}\) aur \(\mathbf{b} \times \mathbf{b} = \mathbf{0}\) use honge.
\[(\mathbf{a}+\mathbf{b}) \times (\mathbf{a}-\mathbf{b}) = \mathbf{a}\times\mathbf{a} - \mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{a} - \mathbf{b}\times\mathbf{b}\]
\[= \mathbf{0} - \mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{a} - \mathbf{0} = \mathbf{b}\times\mathbf{a} + \mathbf{b}\times\mathbf{a} = 2(\mathbf{b} \times \mathbf{a})\]
Answer: Proved.
Question 13: Show that the moment of a force about a point does not depend on where along its line of action the force is applied.
Force \(\mathbf{F}\) ko line of action par point \(A\) se \(A'\) par shift karenge. Position vector mein izafa \(\mathbf{F}\) ki direction mein hoga, aur uska \(\mathbf{F}\) ke sath cross product zero hota hai.
Let the force \(\mathbf{F}\) act at \(A\) with position vector \(\mathbf{r}\). The moment about \(O\) is \(\mathbf{M} = \mathbf{r} \times \mathbf{F}\). Move the application point along the line of action to \(A'\), so \(\mathbf{r}' = \mathbf{r} + t\mathbf{F}\) for some scalar \(t\).
\[\mathbf{M}' = \mathbf{r}' \times \mathbf{F} = (\mathbf{r} + t\mathbf{F}) \times \mathbf{F} = \mathbf{r} \times \mathbf{F} + t(\mathbf{F} \times \mathbf{F}) = \mathbf{r} \times \mathbf{F} + \mathbf{0} = \mathbf{M}\]
Answer: \(\mathbf{M}' = \mathbf{M}\) — the moment is independent of the point of application on the line of action. Proved.
Question 14: A force \(\mathbf{F} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}\) is applied at the point \(A(1, 2, -1)\). Find the moment of the force about the origin.
Moment \(= \mathbf{r} \times \mathbf{F}\), jahan \(\mathbf{r} = \overrightarrow{OA}\) position vector hai.
\[\mathbf{r} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}, \qquad \mathbf{M} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 3 & -2 & 5 \end{vmatrix}\]
\[= (10-2)\mathbf{i} - (5+3)\mathbf{j} + (-2-6)\mathbf{k} = 8\mathbf{i} - 8\mathbf{j} - 8\mathbf{k}\]
Answer: \(8\mathbf{i} - 8\mathbf{j} - 8\mathbf{k}\)
Question 15: A force \(\mathbf{F} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k}\) is applied at \(A(2, -1, 1)\). Find the moment of the force about the point \(M(1, 1, 2)\).
Moment \(= \mathbf{r} \times \mathbf{F}\), jahan \(\mathbf{r} = \overrightarrow{MA} = A - M\).
\[\mathbf{r} = \overrightarrow{MA} = (2-1)\mathbf{i} + (-1-1)\mathbf{j} + (1-2)\mathbf{k} = \mathbf{i} - 2\mathbf{j} - \mathbf{k}\]
\[\mathbf{M} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -1 \\ 2 & 1 & -3 \end{vmatrix} = (6+1)\mathbf{i} - (-3+2)\mathbf{j} + (1+4)\mathbf{k}\]
\[= 7\mathbf{i} + \mathbf{j} + 5\mathbf{k}\]
Answer: \(7\mathbf{i} + \mathbf{j} + 5\mathbf{k}\)
Question 16: A force \(\mathbf{F} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\) acts at the point \(P(3, 1, -2)\). Find the moment of the force about the origin and its magnitude.
Moment \(= \mathbf{r} \times \mathbf{F}\) jahan \(\mathbf{r} = \overrightarrow{OP}\). Phir magnitude nikalenge.
\[\mathbf{M} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -2 \\ 1 & -2 & 2 \end{vmatrix} = (2-4)\mathbf{i} - (6+2)\mathbf{j} + (-6-1)\mathbf{k} = -2\mathbf{i} - 8\mathbf{j} - 7\mathbf{k}\]
\[|\mathbf{M}| = \sqrt{4 + 64 + 49} = \sqrt{117} = 3\sqrt{13}\]
Answer: \(-2\mathbf{i} - 8\mathbf{j} - 7\mathbf{k}\) (magnitude \(3\sqrt{13}\))
