Q70 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

Let $\Pi$ be a plane containing the points $(0,-5,-1),(1,-2,5),(-3,5,0)$ and $L$ be a line passing through the point $(0,-5,-1)$ and parallel to the vector $\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}$. Then the length of the projection of the unit normal vector to the plane $\Pi$ on the line $L$ is

  • A
    $\frac{133 \sqrt{2}}{\sqrt{31}}$
  • Correct answer
    B
    $\frac{14}{\sqrt{682}}$
  • C
    $\frac{133}{\sqrt{31}}$
  • D
    $\frac{268}{2 \sqrt{32}}$

Solution

Equation of plane $(\pi)$ containing the points $(0,-5,-1),(1,-2,5),(-3,5,0)$ is $\left|\begin{array}{ccc}x-0 & y+5 & z+1 \\ 1-0 & -2+5 & 5+1 \\ -3-0 & 5+5 & 0+1\end{array}\right|=0$ $\Rightarrow \quad\left|\begin{array}{ccc}x & y+5 & z+1 \\ 1 & 3 & 6 \\ -3 & 10 & 1\end{array}\right|=0$ $\Rightarrow \quad 3 x+y-z+4=0$ $\Rightarrow \quad \mathbf{r}(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+4=0$ $\Rightarrow \quad \mathbf{r}(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=4$ $\Rightarrow \mathbf{r}\left(\frac{-3}{\sqrt{11}}-\frac{\hat{\mathbf{j}}}{\sqrt{11}}+\frac{\hat{\mathbf{k}}}{\sqrt{11}}\right)=\frac{4}{\sqrt{11}}$ $\therefore$ Unit normal vector to plane $\pi$ is $\hat{\mathbf{n}}=\frac{-3}{\sqrt{11}} \hat{\mathbf{i}}-\frac{\hat{\mathbf{j}}}{\sqrt{11}}+\frac{\hat{\mathbf{k}}}{\sqrt{11}}$ $\therefore$ length of the projection of the unit normal vector $\hat{\mathbf{n}}$ to plane $\pi$ on the line $L$ is $\left|\frac{\left(\frac{-3}{\sqrt{11}} \hat{\mathbf{i}} \frac{-\hat{\mathbf{j}}}{\sqrt{11}} \frac{+\hat{\mathbf{k}}}{\sqrt{11}}\right) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{1+25+36}}\right|$ $=\left|\frac{\frac{-3}{\sqrt{11}} \frac{-5}{\sqrt{11}} \frac{-6}{\sqrt{11}}}{\sqrt{62}}\right|=\left|\frac{-14}{\sqrt{11} \sqrt{62}}\right|=\frac{14}{\sqrt{682}}$

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

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