Let Q be the mirror image of the point P(1,2,1) with respect to the plane x+2y+2z=16.

Question

Let $Q$ be the mirror image of the point $P(1,2,1)$ with respect to the plane $x+2y+2z=16$. Let $T$ be a plane passing through the point $Q$ and containing the line $\vec r=-\hat k+\lambda(\hat i+\hat j+2\hat k)$, $\lambda\in\mathbb R$. Then which of the following points lies on $T$?

Accepted Answer

$P(1,2,1)$ image in plane $x+2 y+2 z=16$

$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2(1+2 \times 2+2 \times 1-16)}{1^{2}+2^{2}+2^{2}} \\
& \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=2 \\
& Q(3,6,5) \\
& \vec{r}=-\hat{k}+\lambda(\hat{i}+\hat{j}+2 \hat{k})
\end{aligned}
$$

$A Q=3 \hat{i}+6 \hat{j}+6 \hat{k}$

$$
\begin{aligned}
& =3(\hat{i}+2 \hat{j}+2 \hat{k}) \\
& \vec{n}=(\hat{i}+2 \hat{j}+2 \hat{k}) \times(\hat{i}+\hat{j}+2 \hat{k}) \\
& \left|\[\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 1 & 1 & 2 \end{array}\end{aligned}$$\right| \\
\& =2 \hat{i}-0 \hat{j}-\hat{k}

\]

Equation of plane $\equiv 2(x-0)+0(y-0)-1(z+1)$ $=0$

$2 x-z=1$

Point lying on plane from the option is $(1,2,1)$ i.e., option (B)

Solutions · 1