The distance of the point having position vector $- \hat{i} + 2 \hat{j} + 6 \hat{k}$ from the straight line passing through the point(2,3,-4) and parallel to the vector, $6 \hat{i} + 3 \hat{j} - 4 \hat{k}$ is :

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$4 \sqrt{3}$

$2 \sqrt{13}$

answer is B.

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Detailed Solution

Let $\vec{n} = 6 \hat{i} + 3 \hat{j} - 4 \hat{k}$

Here, $\vec{A P} = 3 \hat{i} + \hat{j} - 10 \hat{k}$

Now, $A D = |\frac{\vec{A P} \cdot \vec{n}}{| \vec{n} |}|$

$\begin{array}{l} = |\frac{18 + 3 + 40}{\sqrt{36 + 9 + 16}}| = |\frac{61}{\sqrt{61}}| = \sqrt{61} \\ ∴ P D = \sqrt{A P^{2} - A D^{2}} \\ = \sqrt{(9 + 1 + 100) - 61} = \sqrt{110 - 61} = \sqrt{49} = 7 \end{array}$

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