7 Most FAQ’s of Product of Vectors Chapter in Inter 1st Year Maths-1A (TS/AP)

7 Marks

LAQ-1 : If ¯a = 2¯i + ¯j – ¯k, ¯b = -¯i + 2¯j – 4¯k, ¯c = ¯i + ¯j + ¯k then find (¯a x ¯b), (¯b x ¯c)

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

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

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

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

$$= \overline{i}(-4 + 2) – \overline{j}(-8 – 1) + \overline{k}(4 + 1)$$

$$= -2\overline{i} + 9\overline{j} + 5\overline{k} \quad \text{…(1)}$$

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

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

$$= 6\overline{i} – 3\overline{j} – 3\overline{k} \quad \text{…(2)}$$

$$\text{From (1) and (2)}$$

$$(\overline{a} \times \overline{b}) \cdot (\overline{b} \times \overline{c})$$

$$= (-2\overline{i} + 9\overline{j} + 5\overline{k}) \cdot (6\overline{i} – 3\overline{j} – 3\overline{k})$$

$$= (-2)(6) + (9)(-3) + 5(-3)$$

$$= -12 – 27 – 15$$

$$= -54$$

$$(\overline{a} \times \overline{b}) \cdot (\overline{b} \times \overline{c}) = -54$$

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LAQ-2 : If ¯a = 2¯i + ¯j – 3¯k, ¯b = ¯i – 2¯j + ¯k, ¯c = -¯i + ¯j – 4¯k, ¯d = ¯i + ¯j + ¯k, then compute |(¯a x ¯b) x (¯c x ¯d)|

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

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

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

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

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

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

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

$$= 7\overline{i} – 5\overline{j} – 5\overline{k} \quad \text{…(1)}$$

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

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

$$= 5\overline{i} – 3\overline{j} – 2\overline{k} \quad \text{…(2)}$$

$$\text{From (1) and (2)}$$

$$(\overline{a} \times \overline{b}) \times (\overline{c} \times \overline{d}) = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ -5 & -5 & -5 \ 5 & -3 & -2 \end{matrix} \right|$$

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

$$= \overline{i}(10 – 15) – \overline{j}(10 + 25) + \overline{k}(15 + 25)$$

$$= -5\overline{i} – 35\overline{j} + 40\overline{k}$$

$$= 5(-\overline{i} – 7\overline{j} + 8\overline{k})$$

$$|(\overline{a} \times \overline{b}) \times (\overline{c} \times \overline{d})| = 5\sqrt{(-1)^2 + (-7)^2 + 8^2}$$

$$= 5\sqrt{1 + 49 + 64}$$

$$= 5\sqrt{114}$$

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LAQ-3 : If ¯a = ¯i – 2¯j + 3¯k, ¯b = 2¯i + ¯j + ¯k, ¯c = ¯i + ¯j + 2¯k then find |(¯a x ¯b) x ¯c| and |¯a x (¯b x ¯c)|

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

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

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

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

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

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

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

$$= \overline{i}(10 – 5) – \overline{j}(-10 – 5) + \overline{k}(-5 – 5)$$

$$= 5\overline{i} + 15\overline{j} – 10\overline{k}$$

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

$$|(\overline{a} \times \overline{b}) \times \overline{c}| = 5|\overline{i} + 3\overline{j} – 2\overline{k}|$$

$$= 5\sqrt{1^2 + 3^2 + (-2)^2}$$

$$= 5\sqrt{1 + 9 + 4}$$

$$= 5\sqrt{14}$$

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

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

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

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

$$= \overline{i}(-2 + 9) – \overline{j}(1 – 3) + \overline{k}(-3 + 2)$$

$$= 7\overline{i} + 2\overline{j} – \overline{k}$$

$$|\overline{a} \times (\overline{b} \times \overline{c})| = \sqrt{7^2 + 2^2 + (-1)^2}$$

$$= \sqrt{49 + 4 + 1}$$

$$= \sqrt{54}$$

$$= \sqrt{9 \times 6}$$

$$= 3\sqrt{6}$$

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LAQ-4 : If ¯a = ¯i – 2¯j + ¯k, ¯b = 2¯i + ¯j + ¯k, ¯c = ¯i + 2¯j – ¯k then find ¯a x (¯b x ¯c) and |(¯a x ¯b) x ¯c|

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

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

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

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

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

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

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

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

$$= -9\overline{i} – 6\overline{j} – 3\overline{k}$$

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

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

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

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

$$= \overline{i}((-1) – 10) – \overline{j}(3 – 5) + \overline{k}((-6) – 1)$$

$$= -11\overline{i} + 2\overline{j} – 7\overline{k}$$

$$|(\overline{a} \times \overline{b}) \times \overline{c}| = \sqrt{(-11)^2 + 2^2 + (-7)^2}$$

$$= \sqrt{121 + 4 + 49}$$

$$= \sqrt{174}$$

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LAQ-5 : Find the shortest distance between the skew lines ¯r = (6¯i + 2¯j + 2¯k) + t(¯i – 2¯j + 2¯k) and ¯r = (-4¯i – ¯k) + s(3¯i – 2¯j – 2¯k)

$$\text{Given skew lines}$$

$$\overline{r} = (6\overline{i} + 2\overline{j} + 2\overline{k}) + t(\overline{i} – 2\overline{j} + 2\overline{k})$$

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

$$\text{SD} = \frac{|(\overline{a} – \overline{c}) \cdot (\overline{b} \times \overline{d})|}{|\overline{b} \times \overline{d}|}$$

$$\overline{a} = 6\overline{i} + 2\overline{j} + 2\overline{k} \quad \text{and} \quad \overline{b} = \overline{i} – 2\overline{j} + 2\overline{k}$$

$$\overline{c} = -4\overline{i} – \overline{k} \quad \text{and} \quad \overline{d} = 3\overline{i} – 2\overline{j} – 2\overline{k}$$

$$\overline{a} – \overline{c} = (6\overline{i} + 2\overline{j} + 2\overline{k}) – (-4\overline{i} – \overline{k}) = 10\overline{i} + 2\overline{j} + 3\overline{k}$$

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

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

$$= 8\overline{i} + 8\overline{j} + 4\overline{k}$$

$$(\overline{a} – \overline{c}) \cdot (\overline{b} \times \overline{d}) = (10\overline{i} + 2\overline{j} + 3\overline{k}) \cdot (8\overline{i} + 8\overline{j} + 4\overline{k})$$

$$= 10 \times 8 + 2 \times 8 + 3 \times 4$$

$$= 80 + 16 + 12$$

$$= 108$$

$$|\overline{b} \times \overline{d}| = \sqrt{8^2 + 8^2 + 4^2}$$

$$= \sqrt{64 + 64 + 16}$$

$$= \sqrt{144}$$

$$= 12$$

$$\text{SD} = \frac{|(\overline{a} – \overline{c}) \cdot (\overline{b} \times \overline{d})|}{|\overline{b} \times \overline{d}|} = \frac{108}{12} = 9$$

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LAQ-6 : If A = (1, -2, -1), B = (4, 0, – 3), C = (1, 2, -1), D = (2, -4, -5) then find the distance between ¯AB and ¯CD

$$\text{Given points } A = (1, -2, -1)$$

$$B = (4, 0, -3)$$

$$C = (1, 2, -1)$$

$$D = (2, -4, -5)$$

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

$$\overline{OB} = 4\overline{i} – 3\overline{k}$$

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

$$\overline{OD} = 2\overline{i} – 4\overline{j} – 5\overline{k}$$

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

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

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

$$\overline{d} = \overline{CD} = \overline{OD} – \overline{OC} = (2\overline{i} – 4\overline{j} – 5\overline{k}) – (\overline{i} + 2\overline{j} – \overline{k}) = \overline{i} – 6\overline{j} – 4\overline{k}$$

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

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

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

$$= \overline{i}(-8 – 12) – \overline{j}(-12 + 2) + \overline{k}(-18 – 2)$$

$$= -20\overline{i} + 10\overline{j} – 20\overline{k}$$

$$(\overline{a} – \overline{c}) \cdot (\overline{b} \times \overline{d}) = (-4\overline{j}) \cdot (-20\overline{i} + 10\overline{j} – 20\overline{k})$$

$$= (-4)(10)$$

$$= -40$$

$$|\overline{b} \times \overline{d}| = \sqrt{(-20)^2 + 10^2 + (-20)^2}$$

$$= \sqrt{400 + 100 + 400}$$

$$= \sqrt{900}$$

$$= 30$$

$$\text{SD} = \frac{|(\overline{a} – \overline{c}) \cdot (\overline{b} \times \overline{d})|}{|\overline{b} \times \overline{d}|} = \frac{40}{30}$$

$$= \frac{4}{3} \text{ units}$$

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LAQ-7 : If ¯a, ¯b, ¯c are 3 vectors then prove that (¯a x ¯b) x ¯c = (¯c.¯a)¯b – (¯c.¯b) – (¯c.¯b)¯a

$$\text{We take } \overline{a} = a_1\overline{i} + a_2\overline{j} + a_3\overline{k}$$

$$\overline{b} = b_1\overline{i} + b_2\overline{j} + b_3\overline{k}$$

$$\overline{c} = c_1\overline{i} + c_2\overline{j} + c_3\overline{k}$$

$$\overline{a} \times \overline{b} = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{matrix} \right|
= \overline{i}(a_2b_3 – a_3b_2) – \overline{j}(a_1b_3 – a_3b_1) + \overline{k}(a_1b_2 – a_2b_1)$$

$$= \overline{i}(a_2b_3 – a_3b_2) + \overline{j}(a_3b_1 – a_1b_3) + \overline{k}(a_1b_2 – a_2b_1)$$

$$(\overline{a} \times \overline{b}) \times \overline{c} = \left| \begin{matrix} \overline{i} & \overline{j} & \overline{k} \ a_2b_3 – a_3b_2 & a_3b_1 – a_1b_3 & a_1b_2 – a_2b_1 \ c_1 & c_2 & c_3 \end{matrix} \right|$$

$$= \overline{i}[(a_3b_1 – a_1b_3)c_3 – (a_1b_2 – a_2b_1)c_2] – \overline{j}[(a_2b_3 – a_3b_2)c_3 – (a_1b_2 – a_2b_1)c_1] + \overline{k}[(a_2b_3 – a_3b_2)c_2 – (a_3b_1 – a_1b_3)c_1]$$

$$= \overline{i}[a_3b_1c_3 – a_1b_3c_3 – a_1b_2c_2 + a_2b_1c_2] – \overline{j}[a_2b_3c_3 – a_3b_2c_3 – a_1b_2c_1 + a_2b_1c_1] + \overline{k}[a_2b_3c_2 – a_3b_2c_2 – a_3b_1c_1 + a_1b_3c_1] \quad \text{…(1)}$$

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

$$= (c_1a_1 + c_2a_2 + c_3a_3)(b_1\overline{i} + b_2\overline{j} + b_3\overline{k}) – (c_1b_1 + c_2b_2 + c_3b_3)(a_1\overline{i} + a_2\overline{j} + a_3\overline{k})$$

$$= \overline{i}[(c_1a_1 + c_2a_2 + c_3a_3)b_1 – (c_1b_1 + c_2b_2 + c_3b_3)a_1] – \overline{j}[(c_1a_1 + c_2a_2 + c_3a_3)b_2 – (c_1b_1 + c_2b_2 + c_3b_3)a_2] + \overline{k}[(c_1a_1 + c_2a_2 + c_3a_3)b_3 – (c_1b_1 + c_2b_2 + c_3b_3)a_3]$$

$$= \overline{i}[a_1b_1c_1 + a_2b_1c_2 + a_3b_1c_3 – a_1b_1c_3 – a_1b_1c_1 – a_1b_2c_2 – a_1b_3c_3] – \overline{j}[-a_1b_2c_1 – a_2b_2c_2 – a_3b_2c_3 + a_2b_1c_1 + a_2b_2c_2 + a_2b_3c_3] + \overline{k}[a_1b_3c_1 + a_2b_3c_2 + a_3b_3c_3 – a_3b_1c_1 – a_3b_1c_1 – a_3b_2c_2 – a_3b_3c_3]$$

$$= \overline{i}[a_3b_1c_3 – a_1b_3c_3 – a_1b_2c_2 + a_2b_1c_2] – \overline{j}[a_2b_3c_3 – a_3b_2c_3 – a_1b_2c_1 + a_2b_1c_1] + \overline{k}[a_2b_3c_2 – a_3b_2c_2 – a_3b_1c_1 + a_1b_3c_1] \quad \text{…(2)}$$

$$(\overline{a} \times \overline{b}) \times \overline{c} = (\overline{c} \cdot \overline{a})\overline{b} – (\overline{c} \cdot \overline{b})\overline{a}$$