Product Of Vectors (SAQs)

Maths-1A | 5. Product of Vectors – SAQs:
Welcome to SAQs in Chapter 5: Product Of Vectors. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.

Overview:

SAQ-1 : If ¯a = (1, -1, -6), ¯b = (1, -3, 4) and ¯c = (2, -5, 3) then compute ¯a x (¯b x ¯c)

$$\text{Given that } \overline{a} = \overline{i} – \overline{j} – 6\overline{k}$$

$$\overline{b} = \overline{i} – 3\overline{j} + 4\overline{k}$$

$$\overline{c} = 2\overline{i} – 5\overline{j} + 3\overline{k}$$

$$\overline{b} \times \overline{c} = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ 1 & -3 & 4 \ 2 & -5 & 3 \end{matrix} \right|$$

$$= \overline{i}((-3)(3) – (4)(-5)) – \overline{j}((1)(3) – (4)(2)) + \overline{k}((1)(-5) – (-3)(2))$$

$$= \overline{i}(-9 + 20) – \overline{j}(3 – 8) + \overline{k}(-5 + 6)$$

$$= 11\overline{i} + 5\overline{j} + \overline{k}$$

$$\overline{a} \times (\overline{b} \times \overline{c}) = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ 1 & -1 & -6 \ 11 & 5 & 1 \end{matrix} \right|$$

$$= \overline{i}((-1)(1) – (-6)(5)) – \overline{j}((1)(1) – (-6)(11)) + \overline{k}((1)(5) – (-1)(11))$$

$$= \overline{i}(-1 + 30) – \overline{j}(1 + 66) + \overline{k}(5 + 11)$$

$$= 29\overline{i} – 67\overline{j} + 16\overline{k}$$

$$\overline{a} \times (\overline{b} \times \overline{c}) = 29\overline{i} – 67\overline{j} + 16\overline{k}$$

SAQ-2 : If ¯a = 2¯i + 3¯j + 4¯k, ¯b = ¯i + ¯j – ¯k, ¯c = ¯i – ¯j + ¯k, compute ¯a x (¯b x ¯c) and verify that it is perpendicular to ¯a

$$\text{Given } \overline{a} = 2\overline{i} + 3\overline{j} + 4\overline{k}$$

$$\overline{b} = \overline{i} + \overline{j} – \overline{k}$$

$$\overline{c} = \overline{i} – \overline{j} + \overline{k}$$

$$\overline{b} \times \overline{c} = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ 1 & 1 & -1 \ 1 & -1 & 1 \end{matrix} \right|$$

$$= \overline{i}((1)(1) – (-1)(-1)) – \overline{j}((1)(1) – (-1)(1)) + \overline{k}((1)(-1) – (1)(1))$$

$$= \overline{i}(1 – 1) – \overline{j}(1 + 1) + \overline{k}(-1 – 1)$$

$$= -2\overline{j} – 2\overline{k}$$

$$\overline{a} \times (\overline{b} \times \overline{c}) = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ 2 & 3 & 4 \ 0 & -2 & -2 \end{matrix} \right|$$

$$= \overline{i}((3)(-2) – (4)(-2)) – \overline{j}((2)(-2) – (4)(0)) + \overline{k}((2)(-2) – (3)(0))$$

$$= \overline{i}(-6 + 8) – \overline{j}(-4 – 0) + \overline{k}(-4 – 0)$$

$$= 2\overline{i} + 4\overline{j} – 4\overline{k}$$

$$(\overline{a} \times (\overline{b} \times \overline{c})) \cdot \overline{a} = (2\overline{i} + 4\overline{j} – 4\overline{k}) \cdot (2\overline{i} + 3\overline{j} + 4\overline{k})$$

$$= 2(2) + 4(3) – 4(4)$$

$$= 4 + 12 – 16$$

$$= 0$$

SAQ-3 : Find the area of the triangle formed with the points A(1, 2, 3), B(2, 3, 1), C(3, 1, 2)

$$\text{We take } \overline{OA} = \overline{i} + 2\overline{j} + 3\overline{k}$$

$$\overline{OB} = 2\overline{i} + 3\overline{j} + \overline{k}$$

$$\overline{OC} = 3\overline{i} + \overline{j} + 2\overline{k}$$

$$\overline{AB} = \overline{OB} – \overline{OA}$$

$$= (2\overline{i} + 3\overline{j} + \overline{k}) – (\overline{i} + 2\overline{j} + 3\overline{k})$$

$$= \overline{i} + \overline{j} – 2\overline{k}$$

$$\overline{AC} = \overline{OC} – \overline{OA}$$

$$= (3\overline{i} + \overline{j} + 2\overline{k}) – (\overline{i} + 2\overline{j} + 3\overline{k})$$

$$= 2\overline{i} – \overline{j} – \overline{k}$$

$$\overline{AB} \times \overline{AC} = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ 1 & 1 & -2 \ 2 & -1 & -1 \end{matrix} \right|$$

$$= \overline{i}((-1)(-1) – (-2)(-1)) – \overline{j}((1)(-1) – (-2)(2)) + \overline{k}((1)(-1) – (1)(2))$$

$$= \overline{i}(-1 – 2) – \overline{j}(-1 + 4) + \overline{k}(-1 – 2)$$

$$= -3\overline{i} – 3\overline{j} – 3\overline{k}$$

$$|\overline{AB} \times \overline{AC}| = \sqrt{(-3)^2 + (-3)^2 + (-3)^2}$$

$$= \sqrt{9 + 9 + 9}$$

$$= \sqrt{27}$$

$$= \sqrt{9 \times 3}$$

$$= 3\sqrt{3}$$

SAQ-4 : Find unit vector perpendicular to the plane passing through the points (1, 2, 3), (2, -1, 1) and (1, 2, -4)

$$\text{We take } \overline{OA} = \overline{i} + 2\overline{j} + 3\overline{k}$$

$$\overline{OB} = 2\overline{i} – \overline{j} + \overline{k}$$

$$\overline{OC} = \overline{i} + 2\overline{j} – 4\overline{k}$$

$$\overline{AB} = \overline{OB} – \overline{OA}$$

$$= (2\overline{i} – \overline{j} + \overline{k}) – (\overline{i} + 2\overline{j} + 3\overline{k})$$

$$= \overline{i} – 3\overline{j} – 2\overline{k}$$

$$\overline{AC} = \overline{OC} – \overline{OA}$$

$$= (\overline{i} + 2\overline{j} – 4\overline{k}) – (\overline{i} + 2\overline{j} + 3\overline{k})$$

$$= -7\overline{k}$$

$$\overline{AB} \times \overline{AC} = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ 1 & -3 & -2 \ 0 & 0 & -7 \end{matrix} \right|$$

$$= \overline{i}[(-3)(-7) – 0(-2)] – \overline{j}[1(-7) – (0)(-2)] + \overline{k}[1(0) – (0)(-3)]$$

$$= \overline{i}(21) – \overline{j}(-7) + \overline{k}(0)$$

$$= 21\overline{i} + 7\overline{j}$$

$$= 7(3\overline{i} + \overline{j})$$

$$|\overline{AB} \times \overline{AC}| = 7\sqrt{3^2 + 1^2}$$

$$= 7\sqrt{9 + 1}$$

$$= 7\sqrt{10}$$

SAQ-5 : Find the volume of the tetrahedron having the edges ¯i + ¯j + ¯k, ¯i – ¯j, ¯i + 2¯j + ¯k

$$\text{Let } \overline{a} = \overline{i} + \overline{j} + \overline{k}$$

$$\overline{b} = \overline{i} – \overline{j}$$

$$\overline{c} = \overline{i} + 2\overline{j} + \overline{k}$$

$$= \frac{1}{6} \left| 1( (-1)(1) – 0(2)) – 1(1(1) – 0(1)) + 1(1(2) + (-1)(1)) \right|$$

$$= \frac{1}{6} \left| 1(-1) – 1(1) + 1(2 + 1) \right|$$

$$= \frac{1}{6} \left| -1 – 1 + 3 \right|$$

$$= \frac{1}{6} \left| 1 \right|$$

$$= \frac{1}{6} \text{ cubic units}$$

SAQ-6 : Find the volume of the tetrahedron whose vertices are (1, 2, 1), (3, 2, 5), (2, -1, 0), (-1, 0, 1)

$$\text{We take } \overline{OA} = \overline{i} + 2\overline{j} + \overline{k}$$

$$\overline{OB} = 3\overline{i} + 2\overline{j} + 5\overline{k}$$

$$\overline{OC} = 2\overline{i} – \overline{j}$$

$$\overline{OD} = -\overline{i} + \overline{k}$$

$$\overline{AB} = \overline{OB} – \overline{OA}$$

$$= (3\overline{i} + 2\overline{j} + 5\overline{k}) – (\overline{i} + 2\overline{j} + \overline{k})$$

$$= 2\overline{i} + 4\overline{k}$$

$$\overline{AC} = \overline{OC} – \overline{OA}$$

$$= (2\overline{i} – \overline{j}) – (\overline{i} + 2\overline{j} + \overline{k})$$

$$= \overline{i} – 3\overline{j} – \overline{k}$$

$$\overline{AD} = \overline{OD} – \overline{OA}$$

$$= (-\overline{i} + \overline{k}) – (\overline{i} + 2\overline{j} + \overline{k})$$

$$= -2\overline{i} – 2\overline{j}$$

$$\left| \begin{matrix} 2 & 0 & 4 \ 1 & -3 & -1 \ -2 & -2 & 0 \end{matrix} \right|$$

$$= 2 \left( (-3)(0) – (-1)(-2) \right) – 0 \left( (1)(0) – (-2)(4) \right) + 4 \left( (1)(-2) – (-3)(2) \right)$$

$$= 2 \left( 0 – 2 \right) + 4 \left( -2 + 6 \right)$$

$$= 2(-2) + 4(-8)$$

$$= -4 – 32$$

$$= -36$$

SAQ-7 : P.T the smaller angle θ between any two diagonals of a cube is given by cosθ = 1/3

$$\text{Consider a unit cube with vertices } O, A, B, C, L, M, N, P \text{ as shown in the diagram}$$

$$\text{We take } \overline{OA} = \overline{i}, \ \overline{OB} = \overline{j}, \ \overline{OC} = \overline{k}$$

$$\overline{OP} = \overline{OA} + \overline{AL} + \overline{LP} = \overline{i} + \overline{j} + \overline{k} \quad \text{…(1)}$$

$$\overline{CL} = \overline{CN} + \overline{NP} + \overline{PL} = \overline{i} + \overline{j} – \overline{k} \quad \text{…(2)}$$

$$\cos \theta = \frac{\overline{OP} \cdot \overline{CL}}{|\overline{OP}| \ |\overline{CL}|}$$

$$= \frac{(1 \cdot 1) + (1 \cdot 1) + (1 \cdot -1)}{\sqrt{1^2 + 1^2 + 1^2} \ \sqrt{1^2 + 1^2 + (-1)^2}}$$

$$= \frac{1 + 1 – 1}{\sqrt{3} \cdot \sqrt{3}}$$

$$= \frac{1}{3}$$

$$\cos \theta = \frac{1}{3}$$