3D Geometry 2 Question 1
####1. The distance of the point having position vector $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ from the straight line passing through the point $(2,3,-4)$ and parallel to the vector, $6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ is
(a) $2 \sqrt{13}$
(b) $4 \sqrt{3}$
(c) 6
(d) 7
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Answer:
Correct Answer: 1. (d)
Solution:
- Let point $P$ whose position vector is $(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})$ and a straight line passing through $Q(2,3,-4)$ parallel to the vector $\mathbf{n}=6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$.
$\because$ Required distance $d=$ Projection of line segment $P Q$ perpendicular to vector $\mathbf{n}$.
$ =\frac{|\mathbf{P Q} \times \mathbf{n}|}{|\mathbf{n}|} $
Now, $\quad \mathbf{P Q}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-10 \hat{\mathbf{k}}$, so
$ \mathbf{P Q} \times \mathbf{n}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 1 & -10 \\ 6 & 3 & -4 \end{array}\right|=26 \hat{\mathbf{i}}-48 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} $
$ \text { So, } \quad d=\frac{\sqrt{(26)^{2}+(48)^{2}+(3)^{2}}}{\sqrt{(6)^{2}+(3)^{2}+(4)^{2}}} $
$ \begin{aligned} & =\sqrt{\frac{676+2304+9}{36+9+16}}=\sqrt{\frac{2989}{61}} \\ & =\sqrt{49}=7 \text { units } \end{aligned} $
