Let P be the plane passing through the line (x-1)/(1)=(y-2)/(-3)=(z+5)/(7) and the point (…

Question

Let $P$ be the plane passing through the line

$\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$ and the point $(2,4,-3)$.

If the image of the point $(-1,3,4)$ in the plane P

is $(\alpha ,\beta ,\gamma )$ then $\alpha +\beta +\gamma $ is equal to :

Accepted Answer

Equation of line : $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$

Let $B \equiv(2,4,-3)$

So, $\overrightarrow{\mathrm{AB}}=(2-1) \hat{i}+(4-2) \hat{j}+(-3+5) \hat{k}$

$=\hat{i}+2 \hat{j}+2 \hat{k}$

$$
\begin{aligned}
& \vec{n}=\left|\[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 1 & -3 & 7 \ 1 & 2 & 2 \end{array}\end{aligned}$$\right|=(-6-14) \hat{i}-(2-7) \hat{j}+(2+3) \hat{k} \\
\& =-20 \hat{i}+5 \hat{j}+5 \hat{k} \\
\& =-5(4 \hat{i}-\hat{j}-\hat{k})

\]

$\therefore$ Equation of plane is :

$4(x-1)+(-1)(y-2)-1(z+5)=0$

$$
\begin{aligned}
& \Rightarrow 4 x-4-y+2-z-5=0 \\
& \Rightarrow 4 x-y-z-7=0
\end{aligned}
$$

$\because$ Image of point $(-1,3,4)$ is $(\alpha, \beta, \gamma)$

$$
\begin{aligned}
& \text { So, } \frac{\alpha+1}{4}=\frac{\beta-3}{-1}=\frac{\gamma-4}{-1}=\frac{-2(-4-3-4-7)}{16+1+1}=2 \\
& \Rightarrow \alpha=7, \beta=1, \gamma=2 \\
& \text { So, } \alpha+\beta+\gamma=10
\end{aligned}
$$

Solutions · 1