Let a = i ^ + 2 j ^ + 3 k ^ and b = i ^ + j ^ - k ^ . If c is a vector such that a · c = 11 , b · ( a × c ) = 27 and b · c = - 3 ? | b | , then | a × c | 2 is equal to ________ . [2023]

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Solution

Q.
Let $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k} .$ If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = 11,$ $\vec{b} \cdot (\vec{a} \times \vec{c}) = 27$ and $\vec{b} \cdot \vec{c} = - \sqrt{3}$ $| \vec{b} |$ , then $| \vec{a} \times \vec{c} |^{2}$ is equal to ________ . [2023]

Ans.
(285)

$Given, \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k},$ $\vec{b} = \hat{i} + \hat{j} - \hat{k}$

$∴$ $\vec{a} \cdot \vec{b} = 1 + 2 - 3 = 0$

$Now, \vec{b} \times (\vec{a} \times \vec{c}) = (\vec{b} \cdot \vec{c}) \vec{a} - (\vec{b} \cdot \vec{a}) \vec{c} = (\vec{b} \cdot \vec{c}) \vec{a} = - \sqrt{3}$ $| \vec{b} |$ $\vec{a}$

$= - \sqrt{3} (\sqrt{1 + 1 + 1}) \vec{a} = - \sqrt{3} (\sqrt{3}) \vec{a} = - 3 \vec{a}$

Let $θ$ be the angle between $\vec{b}$ and $\vec{a} \times \vec{c}$ .

$∴$ $| \vec{b} \times (\vec{a} \times \vec{c}) |$ $=$ $| \vec{b} |$ $\cdot$ $| \vec{a} \times \vec{c} |$ $\sin θ$

$\Rightarrow$ $| - 3 \vec{a} |$ $=$ $| \vec{b} |$ $\cdot$ $| \vec{a} \times \vec{c} |$ $\sin θ$

$∴$ $| \vec{b} |$ $\cdot$ $| \vec{a} \times \vec{c} |$ $\sin θ = 3 \sqrt{14}$ ...(i)

Also, $\vec{b} \cdot (\vec{a} \times \vec{c}) = 27 (Given)$

$∴$ $| \vec{b} |$ $\cdot$ $| \vec{a} \times \vec{c} |$ $\cos θ = 27$ ...(ii)

Dividing (i) by (ii), we get

$\tan θ = \frac{3 \sqrt{14}}{27}$

$\Rightarrow \sec^{2} θ = 1 + \tan^{2} θ = 1 + \frac{9 \times 14}{{(27)}^{2}} = 1 + \frac{14}{81} = \frac{95}{81}$

$\Rightarrow \cos^{2} θ = \frac{81}{95} \Rightarrow \sin^{2} θ = 1 - \frac{81}{95} = \frac{14}{95} \Rightarrow \sin θ = \sqrt{\frac{14}{95}}$

From (i), we get

$\sqrt{3}$ $| \vec{a} \times \vec{c} |$ $= 3 \sqrt{14} \times \frac{\sqrt{95}}{\sqrt{14}}$ $\Rightarrow$ $| \vec{a} \times \vec{c} |$ $= \sqrt{3 \times 95}$

$∴$ $| \vec{a} \times \vec{c} |^{2}$ $= 3 \times 95 = 285$

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