JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 9)
Let $$\overrightarrow{\mathrm{a}}=6 \hat{i}+\hat{j}-\hat{k}$$ and $$\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j}$$. If $$\overrightarrow{\mathrm{c}}$$ is a is vector such that $$|\overrightarrow{\mathrm{c}}| \geq 6, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=6|\overrightarrow{\mathrm{c}}|,|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}$$ and the angle between $$\vec{a} \times \vec{b}$$ and $$\vec{c}$$ is $$60^{\circ}$$, then $$|(\vec{a} \times \vec{b}) \times \vec{c}|$$ is equal to:
$$\frac{3}{2} \sqrt{6}$$
$$\frac{9}{2}(6-\sqrt{6})$$
$$\frac{9}{2}(6+\sqrt{6})$$
$$\frac{3}{2} \sqrt{3}$$
Explanation
$$\begin{aligned}
& |(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 60^{\circ} \\
& \left|\begin{array}{ccc}
i & j & k \\
6 & 1 & -1 \\
1 & 1 & 0
\end{array}\right|=i(1)-j(1)+k(5) \\
& =i-j+5 k \\
& |\vec{a} \times \vec{b}|=\sqrt{1+1+25}=\sqrt{27} \\
& |\vec{c}-\vec{a}|=2 \sqrt{2} \\
& c^2+a^2-2 a c=8 \\
& c^2-12 c+30=0 \\
& c=\frac{12+\sqrt{24}}{2}=6+\sqrt{6} \\
& \Rightarrow|(\vec{a} \times \vec{b}) \times \vec{c}|=\sqrt{27} \times(6+\sqrt{6}) \times \frac{\sqrt{3}}{2} \\
& =\frac{a}{2}(6+\sqrt{6}) \\
\end{aligned}$$
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