3D Geometry Ques 2
- The length of the projection of the line segment joining the points $(5,-1,4)$ and $(4,-1,3)$ on the plane, $x+y+z=7$ is
(2018 Main)
(a) $\frac{2}{\sqrt{3}}$
(b) $\frac{2}{3}$
(c) $\frac{1}{3}$
(d) $\sqrt{\frac{2}{3}}$
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Answer:
Correct Answer: 2.(d)
Solution: (d) Key Idea length of projection of the line segment joining $\mathrm{a}_1$ and $\mathrm{a}_2$ on the plane $\mathrm{r} \cdot \mathrm{n}=d$ is $\left|\frac{\left(\mathbf{a} _{\mathbf{2}}-\mathbf{a} _{\mathbf{1}}\right) \times \mathbf{n}}{|\mathbf{n}|}\right|$
Length of projection the line segment joining the points $(5,-1,4)$ and $(4,-1,3)$ on the plane $x+y+z=7$ is
$ \begin{aligned} A C=&\left|\frac{\left(\mathbf{a}_2-\mathbf{a}_1\right) \times \mathbf{n}}{|\mathbf{n}|}\right| =\frac{|(-\hat{\mathrm{i}}-\hat{\mathrm{k}}) \times(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})|}{|\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}|} \\ A C =&\frac{|\hat{\mathrm{i}}-\hat{\mathrm{k}}|}{\sqrt{3}} \Rightarrow A C=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{\frac{2}{3}} \end{aligned} $
Alternative Method
Clearly, DR’s of $A B$ are $4-5,-1+1,3-4$, i.e. $-1,0,-1$ and DR’s of normal to plane are $1,1,1$.
Now, let $\theta$ be the angle between the line and plane, then $\theta$ is given by $\sin \theta=\frac{|-1+0-1|}{\sqrt{(-1)^2+(-1)^2} \sqrt{1^2+1^2+1^2}}$
$=\frac{2}{\sqrt{2} \sqrt{3}}=\sqrt{\frac{2}{3}}$
$ \Rightarrow \sin \theta=\sqrt{\frac{2}{3}} \Rightarrow \cos \theta=\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}$
Clearly, length of projection $ =A B \cos \theta=\sqrt{2} \frac{1}{\sqrt{3}} $ $\quad[\because \mathrm{AB}=\sqrt{2}] $
$= \sqrt{\frac{2}{3}}$
BETA
