the perpendicular distance from the origin to
Question:

the perpendicular distance from the origin to the plane containing the two lines,

$\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}$ and $\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}$ is

  1. $11 / \sqrt{6}$

  2. $6 \sqrt{11}$

  3. 11

  4. $11 \sqrt{6}$


Correct Option: 1

Solution:

$\left|\begin{array}{lll}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7\end{array}\right|$

$\hat{\mathrm{i}}(35-28)-\hat{\mathrm{j}}(21.7)+\hat{\mathrm{k}}(12-5)$

$7 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$

$\hat{i}-2 \hat{j}+\hat{k}$

$1(x+2)-2(y-2)+1(z+15)=0$

$x-2 y+z+11=0$

$\frac{11}{\sqrt{4+1+1}}=\frac{11}{\sqrt{6}}$

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