Q100 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
The foot of the perpendicular drawn from $A(1,2,2)$ onto the plane $x+2 y+2 z-5=0$ is $B(\alpha, \beta, \gamma)$. If $\pi(x, y, z) \equiv x+2 y+2 z+5=0$ is a plane then $-\pi(A):$ $\pi(B)=$
- A$15: 32$
- Correct answerB$-7: 5$
- C$-15: 47$
- D$-27: 20$
Solution
$\overrightarrow{\mathrm{AB}}=(\alpha-1) \hat{i}+(\beta-2) \hat{j}+(\gamma-2) \hat{k}$
Normal vector of plane $x+2 y+2 z-5=0$ $\overrightarrow{\mathrm{N}}=\hat{i}+2 \hat{j}+2 \hat{k}$
As $\overrightarrow{\mathrm{AB}} \| \overrightarrow{\mathrm{N}}$, So, $\frac{\alpha-1}{1}=\frac{\beta-2}{2}=\frac{\gamma-2}{2}=\lambda$ $\alpha=\lambda+1, \beta=2 \lambda+2, \gamma=2 \lambda+2$
Also, $(\alpha, \beta, \gamma)$ lies on $x+2 y+2 z-5=0$ $\begin{aligned} & \therefore \lambda+1+2(2 \lambda+2)+2(2 \lambda+2)-5=0 \\ & \Rightarrow \lambda=\frac{-4}{9} \therefore \mathrm{~B}(\alpha, \beta, \gamma)=\left(\frac{5}{9}, \frac{1}{9}, \frac{1}{9}\right) \\ & -\pi(\mathrm{A}): \pi(\mathrm{B})=-7: 5\end{aligned}$
Normal vector of plane $x+2 y+2 z-5=0$ $\overrightarrow{\mathrm{N}}=\hat{i}+2 \hat{j}+2 \hat{k}$
As $\overrightarrow{\mathrm{AB}} \| \overrightarrow{\mathrm{N}}$, So, $\frac{\alpha-1}{1}=\frac{\beta-2}{2}=\frac{\gamma-2}{2}=\lambda$ $\alpha=\lambda+1, \beta=2 \lambda+2, \gamma=2 \lambda+2$
Also, $(\alpha, \beta, \gamma)$ lies on $x+2 y+2 z-5=0$ $\begin{aligned} & \therefore \lambda+1+2(2 \lambda+2)+2(2 \lambda+2)-5=0 \\ & \Rightarrow \lambda=\frac{-4}{9} \therefore \mathrm{~B}(\alpha, \beta, \gamma)=\left(\frac{5}{9}, \frac{1}{9}, \frac{1}{9}\right) \\ & -\pi(\mathrm{A}): \pi(\mathrm{B})=-7: 5\end{aligned}$
Hence, the correct answer is option: 'B'.