Let and . If is a vector such that and the angle between and is 60°, then is equal to : [2024]

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Solution

Q.
Let $\vec{a} = 6 \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$ . If $\vec{c}$ is a vector such that $| \vec{c} | \geq 6, \vec{a} \cdot \vec{c} = 6 | \vec{c} |, | \vec{c} - \vec{a} | = 2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is 60°, then $| (\vec{a} \times \vec{b}) \times \vec{c} |$ is equal to : [2024]

1 $\frac{3}{2} \sqrt{3}$

2 $\frac{9}{2} (6 - \sqrt{6})$

3 $\frac{9}{2} (6 + \sqrt{6})$

4 $\frac{3}{2} \sqrt{6}$

Ans.
(3)

We have, $| (\vec{a} \times \vec{b}) \times \vec{c} |$ $= | \vec{a} \times \vec{b} | | \vec{c} | \sin 60 °$

Now, $\vec{a} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 1 & - 1 \\ 1 & 1 & 0 \end{matrix} |$ $= \hat{i} - \hat{j} + 5 \hat{k}$

So, $| \vec{a} \times \vec{b} |$ $= \sqrt{1 + 1 + 25} = \sqrt{27} = 3 \sqrt{3}$

Also, $| \vec{c} - \vec{a} |$ $= 2 \sqrt{2}$ [Given]

$\Rightarrow | \vec{c} |^{2} + | \vec{a} |^{2} - 2 | \vec{c} \cdot \vec{a} | = 8$

$\Rightarrow | \vec{c} |^{2} + 38 - 12 | \vec{c} | = 8$ $[∵ \vec{c} \cdot \vec{a} = 6 | \vec{c} |]$

$\Rightarrow | \vec{c} |^{2} - 12 | \vec{c} | + 30 = 0$

$\Rightarrow | \vec{c} | = \frac{12 + \sqrt{24}}{2}$ $[∵ | \vec{c} | \geq 6]$

$\Rightarrow | \vec{c} | = 6 + \sqrt{6}$

$∴$ $| (\vec{a} \times \vec{b}) \times \vec{c} |$ $= \sqrt{27} \times (6 + \sqrt{6}) \times \frac{\sqrt{3}}{2} = \frac{9}{2} (6 + \sqrt{6})$ .

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