Let three vectors , form a triangle such that and the area of the triangle is . If is a positive real number, then is equal to: [2024]

Question

Let three vectors $\vec{a} = α \hat{i} + 4 \hat{j} + 2 \hat{k}$ , $\vec{b} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k},$ $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$ form a triangle such that $\vec{c} = \vec{a} - \vec{b}$ and the area of the triangle is $5 \sqrt{6}$ . If $α$ is a positive real number, then $| \vec{c} |^{2}$ is equal to: [2024]

Smart Achievers · Accepted Answer

(1)

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

$\Rightarrow \vec{c} = (α - 5) \hat{i} + \hat{j} - 2 \hat{k}$

Let ABC be the given triangle.

Area of $△ A B C = \frac{1}{2} | \vec{a} \times \vec{b} | = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ α & 4 & 2 \\ 5 & 3 & 4 \end{matrix} | |$

$\Rightarrow \frac{1}{2}$ $| 10 \hat{i} - (4 α - 10) \hat{j} + (3 α - 20) \hat{k} |$ $= 5 \sqrt{6}$ [Given]

$= 100 + {(4 α - 10)}^{2} + {(3 α - 20)}^{2} = 600$

$\Rightarrow 25 α^{2} - 200 α = 0 \Rightarrow α = 8$ $[∵ α .]$

$\Rightarrow | \vec{c} |^{2} = 9 + 1 + 4 = 14$ .

Solution

1 14

2 16

3 12

4 10

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