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Question:
The length of the perpendicular drawn from the point $(2,1,4)$ to the plane containing the lines
$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})$ and
$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})+\mu(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})$ is :
Correct Option: 1
Solution:
perpendicular vector to the plane
$\overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 2 & -1 \\ -1 & 1 & -2\end{array}\right|=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
Eq. of plane
$-3(x-1)+3(y-1)+3 z=0$
$\Rightarrow x-y-z=0$
$\mathrm{d}_{(2,1,4)}=\frac{|2-1-4|}{\sqrt{1^{2}+1^{2}+1^{2}}}=\sqrt{3}$
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