Let , and be three vectors such that . If the angle between the vector and the vector is , then the greatest integer less than or equal to is __________. [2024]

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Solution

Q.
Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = - \hat{i} - 8 \hat{j} + 2 \hat{k}$ and $\vec{c} = 4 \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ be three vectors such that $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$ . If the angle between the vector $\vec{c}$ and the vector $3 \hat{i} + 4 \hat{j} + \hat{k}$ is $θ$ , then the greatest integer less than or equal to $\tan^{2} θ$ is __________. [2024]

Ans.
(38)

We have, $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$

$\Rightarrow - 10 \hat{i} + 3 \hat{j} + 7 \hat{k} = \hat{i} (c_{2} - c_{3}) - \hat{j} (4 - c_{3}) + \hat{k} (4 - c_{2})$

On comparing, we get $c_{2} = - 3, c_{3} = 7$

$∴ \vec{c} = 4 \hat{i} - 3 \hat{j} + 7 \hat{k}$

Now, $\frac{\vec{c} \cdot (3 \hat{i} + 4 \hat{j} + \hat{k})}{| \vec{c} | | 3 \hat{i} + 4 \hat{j} + \hat{k} |} \Rightarrow \cos θ$

$\Rightarrow \cos θ = \frac{12 - 12 + 7}{\sqrt{74} \sqrt{26}} \Rightarrow \cos θ = \frac{7}{2 \sqrt{481}}$

So, $\tan θ = \frac{25 \sqrt{3}}{7} \Rightarrow \tan^{2} θ = 38.26$

$\Rightarrow [\tan^{2} θ] = [38.26] = 38$

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