Q72 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics

If $\overrightarrow{\mathrm{p}}=4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ is a point and $\overrightarrow{\mathrm{q}}=9 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ is a vector, then the perpendicular distance of origin from the plane passing through $\overrightarrow{\mathrm{p}}$ and perpendicular to $\overrightarrow{\mathrm{q}}$ is

  • Correct answer
    A
    4
  • B
    $3 \sqrt{2}$
  • C
    9
  • D
    11

Solution

$\vec{p}=4 \hat{i}-\hat{j}-\hat{k} ; \vec{q}=9 \hat{i}-2 \hat{j}-6 \hat{k}$ Equation of plane passing through $\vec{p}$ and perpendicular to $\vec{q}$ is $\vec{q} \cdot(\vec{r}-\vec{p})=0$ $\vec{q} \cdot \vec{r}-\vec{q} \cdot \vec{p}=0$ $\begin{array}{r}\Rightarrow(9 \hat{i}-2 \hat{j}+6 \hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})-(9 \hat{i}-2 \hat{j}+6 \hat{k}) \\ (9 \hat{i}-\hat{j}+\hat{k})=0\end{array}$ $\Rightarrow 9 x-2 y+6 z-44=0$ Now, the perpendicular distance from origin $=\left|\frac{-44}{\sqrt{81+4+36}}\right|=\left|\frac{-44}{\sqrt{121}}\right|=\frac{44}{11}=4$

Hence, the correct answer is option: 'A'.

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