Q73 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
$\vec{r}$ is a vector perpendicular to the piane determined by the vectors $2 \hat{i}-\hat{j}$ and $\hat{j}+2 \hat{k}$. If the magnitude of the projection of $\vec{r}$ on the vector $2 \hat{i}+\hat{j}+2 \hat{k}$ is 1 , then $|\vec{r}|=$
- A$\sqrt{6}$
- B$3 \sqrt{6}$
- C$\frac{2 \sqrt{6}}{3}$
- Correct answerD$\frac{3 \sqrt{6}}{2}$
Solution
Normal of plane
$\vec{r}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 0 \\
0 & 1 & 2
\end{array}\right|=-2 \hat{i}-4 \hat{j}+2 \hat{k}$
Since, $\vec{r}$ perpendicular to plane $\therefore \vec{r}=\lambda(2 \hat{i}+4 \hat{j}-2 \hat{k})$ Projection of $\vec{r} \text { on } 2 \hat{i}+\hat{j}+2 \hat{k}$ $\therefore \frac{\lambda(2 \hat{i}+4 \hat{j}-2 \hat{k}) \cdot(2 \hat{i}+\hat{j}+2 \hat{k})}{\sqrt{4+1+4}}=1$ $\Rightarrow \frac{\lambda(4+4-4)}{3}=1 \Rightarrow \lambda=\frac{3}{4}$ $\therefore|\vec{r}|=\frac{3}{4} \sqrt{4+16+4}=\frac{3}{4} \cdot 2 \sqrt{6}=\frac{3 \sqrt{6}}{2}$
Since, $\vec{r}$ perpendicular to plane $\therefore \vec{r}=\lambda(2 \hat{i}+4 \hat{j}-2 \hat{k})$ Projection of $\vec{r} \text { on } 2 \hat{i}+\hat{j}+2 \hat{k}$ $\therefore \frac{\lambda(2 \hat{i}+4 \hat{j}-2 \hat{k}) \cdot(2 \hat{i}+\hat{j}+2 \hat{k})}{\sqrt{4+1+4}}=1$ $\Rightarrow \frac{\lambda(4+4-4)}{3}=1 \Rightarrow \lambda=\frac{3}{4}$ $\therefore|\vec{r}|=\frac{3}{4} \sqrt{4+16+4}=\frac{3}{4} \cdot 2 \sqrt{6}=\frac{3 \sqrt{6}}{2}$
Hence, the correct answer is option: 'D'.