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Let $\mathrm{Q}$ be the mirror image of the point $\mathrm{P(1,2,1)}$ with respect to the plane $\mathrm{x+2y+2z=16}$ . Let $\mathrm{T}$ be a plane passing through the point $\mathrm{Q}$ and contains 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 $\mathrm{T}$
Option: 1
$(2,1,0)$

Option: 2
$(1,2,1)$

Option: 3
$(1,2,2)$

Option: 4
$(1,3,2)$

Answers (1)

best_answer

Let coordinate of $\mathrm{Q}$ in $\mathrm{x+2 y+2 z=16}$ be $\mathrm{\left(x_{1}, y_{1}, z_{1}\right)}$

$\mathrm{\therefore \frac{x_{1}-1}{1}=\frac{y_{1}-2}{2}=\frac{z_{1}-1}{2}=\frac{-2(1+4+2-16)}{1+2^{2}+2^{2}}} \\$

$\mathrm{\Rightarrow \frac{x_{1}-1}{1}=\frac{y_{1}-2}{2}=\frac{z_{1}-1}{2}=2} \\$

$\mathrm{\Rightarrow x_{1}=3, \quad y_{1}=6,\quad z_{1}=5 }$

Vector $\vec{b}$ parallel to given line is $\mathrm{i+j+2 k}$

$\mathrm{R(0,0,-1)}$ lies on line, so $\mathrm{\overrightarrow{R Q}}$ is also parallel to plane

$\mathrm{\overrightarrow{R Q}=3 i+6 j+6 k}$

$\mathrm{\overrightarrow{n}}$ of plane $\mathrm{=(i+j+2 k) \times(3 i+6 j+6 k)}$

$\mathrm{=\left|\begin{array}{ccc} i & j & k \\ 1 & 1 & 2 \\ 3 & 6 & 6 \end{array}\right|} \\$

$\mathrm{=-6 i-j(0)+k(3)} \\$

$\mathrm{=-6 i+3 k}$

Equation of plane

$\mathrm{-6(x-0)+3(z+1)=0} \\$

$\mathrm{\Rightarrow -6 x+3 z+3=0} \\$

$\mathrm{\Rightarrow -2 x+z+1=0}$

Checking options,option 2 lies on this plane

Hence answer is option 2

Posted by

Gautam harsolia

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