Product Of Vectors (VSAQs)

Maths-1A | 5. Product of Vectors – VSAQs:
Welcome to VSAQs in Chapter 5: Product Of Vectors. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : If ¯a = ¯i + 2¯j – 3k, ¯b = 3¯i – ¯j + 2¯k then S.T ¯a + ¯b, ¯a – ¯b are perpendicular

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

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

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

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

$$(\overline{a} + \overline{b}) \cdot (\overline{a} – \overline{b})$$

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

$$= 4(-2) + 1(3) + (-1)(-5)$$

$$= -8 + 3 + 5$$

$$= -8 + 8$$

$$= 0$$

$$(\overline{a} + \overline{b}), (\overline{a} – \overline{b}) \text{ are perpendicular.}$$

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VSAQ-2 : If |¯a + ¯b| = |¯a – ¯b| then find the angle between ¯a and ¯b

$$\text{Given that } |\overline{a} + \overline{b}| = |\overline{a} – \overline{b}|$$

$$|\overline{a} + \overline{b}|^2 = |\overline{a} – \overline{b}|^2$$

$$\overline{a}^2 + \overline{b}^2 + 2\overline{a} \cdot \overline{b} = \overline{a}^2 + \overline{b}^2 – 2\overline{a} \cdot \overline{b}$$

$$4\overline{a} \cdot \overline{b} = 0$$

$$\overline{a} \cdot \overline{b} = 0$$

$$\text{Angle between } \overline{a} \text{ and } \overline{b} \text{ is } 90^\circ$$

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VSAQ-3 : If the vectors 2¯i + λ¯j – ¯k and 4¯i – 2¯j + 2¯k are perpendicular to each other then find λ

$$\text{Let } \overline{a} = 2\overline{i} + \lambda\overline{j} – \overline{k}$$

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

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

$$2(4) + \lambda(-2) + (-1)(2) = 0$$

$$8 – 2\lambda – 2 = 0$$

$$6 – 2\lambda = 0$$

$$2\lambda = 6$$

$$\lambda = 3$$

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VSAQ-4 : If the vectors λ¯i – 3¯j + 5¯k, 2λ¯i – λ¯j – ¯k are perpendicular to each other find λ

$$\text{Let } \overline{a} = \lambda\overline{i} – 3\overline{j} + 5\overline{k}$$

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

$$(\lambda\overline{i} – 3\overline{j} + 5\overline{k}) \cdot (2\lambda\overline{i} – \lambda\overline{j} – \overline{k}) = 0$$

$$\lambda(2\lambda) + (-3)(-\lambda) + 5(-1) = 0$$

$$2\lambda^2 + 3\lambda – 5 = 0$$

$$( \lambda – 1 )( 2\lambda + 5 ) = 0$$

$$\lambda – 1 = 0$$

$$\lambda = 1$$

$$2\lambda + 5 = 0$$

$$2\lambda = -5$$

$$\lambda = -\frac{5}{2}$$

$$\lambda = 1 \text{ or } -\frac{5}{2}$$

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VSAQ-5 : Find the angle between the vectors ¯i + 2¯j + 3¯k and 3¯i – ¯j + 2¯k

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

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

$$\cos \theta = \frac{(\overline{i} + 2\overline{j} + 3\overline{k}) \cdot (3\overline{i} – \overline{j} + 2\overline{k})}{\sqrt{1^2 + 2^2 + 3^2} \sqrt{3^2 + (-1)^2 + 2^2}}$$

$$= \frac{1(3) + 2(-1) + 3(2)}{\sqrt{1 + 4 + 9} \sqrt{9 + 1 + 4}}$$

$$= \frac{3 – 2 + 6}{\sqrt{14} \sqrt{14}}$$

$$= \frac{7}{14}$$

$$= \frac{1}{2} = \cos 60^\circ$$

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VSAQ-6 : Find the angle between the planes ¯r.(2¯i – ¯j + 2¯k) = 3, ¯r.(3¯i + 6¯j + ¯k) = 4

$$\text{Comparing the given planes with}$$

$$\overline{r} \cdot \overline{n_1} = p_1, \quad \overline{r} \cdot \overline{n_2} = p_2 \quad \text{we get}$$

$$\overline{n_1} = 2\overline{i} – \overline{j} + 2\overline{k}$$

$$\overline{n_2} = 3\overline{i} + 6\overline{j} + \overline{k}$$

$$\cos \theta = \frac{(2\overline{i} – \overline{j} + 2\overline{k}) \cdot (3\overline{i} + 6\overline{j} + \overline{k})}{\sqrt{4 + 1 + 4} \sqrt{9 + 36 + 1}}$$

$$= \frac{2(3) – 1(6) + 2(1)}{\sqrt{9} \sqrt{46}}$$

$$= \frac{6 – 6 + 2}{3 \sqrt{46}}$$

$$= \frac{2}{3 \sqrt{46}}$$

$$\cos \theta = \frac{2}{3 \sqrt{46}}$$

$$\theta = \cos^{-1} \left( \frac{2}{3 \sqrt{46}} \right)$$

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VSAQ-7 : Find the area of the parallelogram whose adjacent sides are ¯a = 2¯j – ¯k, ¯b = -¯i + ¯k

$$\text{Given sides } \overline{a} = 2\overline{j} – \overline{k}$$

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

$$\overline{a} \times \overline{b} = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ 0 & 2 & -1 \ -1 & 0 & 1 \end{matrix} \right|$$

$$= \overline{i} \left( 2(1) – (-1)(0) \right) – \overline{j} \left( 0(1) – (-1)(-1) \right) + \overline{k} \left( 0(0) – (-1)(2) \right)$$

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

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

$$|\overline{a} \times \overline{b}| = \sqrt{(2)^2 + (1)^2 + (2)^2}$$

$$= \sqrt{4 + 1 + 4}$$

$$= \sqrt{9}$$

$$= 3$$

$$= 3 \text{ Sq.units}$$

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VSAQ-8 : Find the area of the parallelogram whose diagonals are 3¯i + ¯j – 2¯k, ¯i – 3¯j + 4¯k

$$\text{Given diagonals } \overline{d_1} = 3\overline{i} + \overline{j} – 2\overline{k}$$

$$\overline{d_2} = \overline{i} – 3\overline{j} + 4\overline{k}$$

$$\overline{d_1} \times \overline{d_2} = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ 3 & 1 & -2 \ 1 & -3 & 4 \end{matrix} \right|$$

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

$$= \overline{i} (4 – 6) – \overline{j} (12 + 2) + \overline{k} (-9 – 1)$$

$$= -2\overline{i} – 14\overline{j} – 10\overline{k}$$

$$= -2(\overline{i} + 7\overline{j} + 5\overline{k})$$

$$|\overline{d_1} \times \overline{d_2}| = |-2| \sqrt{(1)^2 + (7)^2 + (5)^2}$$

$$= 2 \sqrt{1 + 49 + 25}$$

$$= 2 \sqrt{75}$$

$$\text{Area} = \frac{1}{2} |\overline{d_1} \times \overline{d_2}|$$

$$= \frac{1}{2} \cdot 2 \sqrt{75}$$

$$= \sqrt{75}$$

$$= \sqrt{25 \times 3}$$

$$= 5\sqrt{3} \text{ Sq.units}$$

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VSAQ- 9 : Find a unit vector perpendicular to the plane containing the vectors ¯a = 4¯i + 3¯j – ¯k, ¯b = 2¯i – 6¯j – 3¯k

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

$$\overline{b} = 2\overline{i} – 6\overline{j} – 3\overline{k}$$

$$\overline{a} \times \overline{b} = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ 4 & 3 & -1 \ 2 & -6 & -3 \end{matrix} \right|$$

$$= \overline{i} \left( (3)(-3) – (-6)(-1) \right) – \overline{j} \left( (4)(-3) – (2)(-1) \right) + \overline{k} \left( (4)(-6) – (3)(2) \right)$$

$$= \overline{i} (-9 – 6) – \overline{j} (-12 + 2) + \overline{k} (-24 – 6)$$

$$= -15\overline{i} + 10\overline{j} – 30\overline{k}$$

$$= 5(-3\overline{i} + 2\overline{j} – 6\overline{k})$$

$$|\overline{a} \times \overline{b}| = 5 \sqrt{(-3)^2 + (2)^2 + (-6)^2}$$

$$= 5 \sqrt{9 + 4 + 36}$$

$$= 5 \sqrt{49}$$

$$= 5 \times 7$$

$$= 35$$

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VSAQ-10 : Find the area of the triangle having (3¯i + 4¯j), (-5¯i + 7¯j) as adjacent sides

$$\text{Let } \overline{a} = 3\overline{i} + 4\overline{j}$$

$$\overline{b} = -5\overline{i} + 7\overline{j}$$

$$\overline{a} \times \overline{b} = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ 3 & 4 & 0 \ -5 & 7 & 0 \end{matrix} \right|$$

$$= \overline{i}(0 – 0) – \overline{j}(0 – 0) + \overline{k}(3 \cdot 7 – 4 \cdot (-5))$$

$$= \overline{i}(0) – \overline{j}(0) + \overline{k}(21 + 20)$$

$$= 41\overline{k}$$

$$|\overline{a} \times \overline{b}| = |41\overline{k}|$$

$$|\overline{a} \times \overline{b}| = 41$$

$$\text{Area of the triangle} = \frac{1}{2} |\overline{a} \times \overline{b}|$$

$$= \frac{1}{2} \times 41$$

$$= 20.5 \text{ sq.units}$$