If for a>0, the feet of perpendiculars

Question:

If for $\mathrm{a}>0$, the feet of perpendiculars from the points $\mathrm{A}(\mathrm{a},-2 \mathrm{a}, 3)$ and $\mathrm{B}(0,4,5)$ on the plane $l_{\mathrm{X}}+\mathrm{my}+\mathrm{nz}=0$ are points $\mathrm{C}(0,-\mathrm{a},-1)$ and $\mathrm{D}$ respectively, then the length of line segment $C D$ is equal to :

  1. (1) $\sqrt{31}$

  2. (2) $\sqrt{41}$

  3. (3) $\sqrt{55}$

  4. (4) $\sqrt{66}$


Correct Option: , 4

Solution:

C lies on plane $\Rightarrow-m a-n=0 \Rightarrow \frac{m}{n}=-\frac{1}{a} \ldots$ (1)

$\overrightarrow{\mathrm{CA}} \| \hat{i}+m \hat{j}+n \hat{k}$

$\frac{a-0}{l}=\frac{-a}{m}=\frac{4}{n} \Rightarrow \frac{m}{n}=-\frac{a}{4}$

From (1) \& (2)

$-\frac{1}{a}=\frac{-a}{4} \Rightarrow a^{2}=4 \Rightarrow a=2 \quad($ since $a>0)$

Let $m=-t \Rightarrow n=2 t$

$\frac{2}{l}=\frac{-2}{-t} \Rightarrow l=\mathrm{t}$

$\frac{2}{l}=\frac{-2}{-\mathrm{t}} \Rightarrow l=\mathrm{t}$

So plane : $t(x-y+2 z)=0$

$\mathrm{BD}=\frac{6}{\sqrt{6}}=\sqrt{6} \quad \mathrm{C} \cong(0,-2,-1)$

$\mathrm{CD}=\sqrt{\mathrm{BC}^{2}-\mathrm{BD}^{2}}$

$=\sqrt{\left(0^{2}+6^{2}+6^{2}\right)-(\sqrt{6})^{2}}$

$=\sqrt{66}$

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