If the image of the point P(1,0,3) in the line joining the points A(4,7,1) and B(3,5,3) is…

Question

If the image of the point $P(1,0,3)$ in the line joining the points $A(4,7,1)$ and $B(3,5,3)$ is $Q(\alpha ,\beta ,\gamma )$, then $\alpha +\beta +\gamma $ is equal to :

Accepted Answer

Let $X$ be mid point of $P$ and $Q$, which would be also feet of perpendicular.

Let $X$ divides $A$ and $B$ in $\lambda: 1, \lambda \neq-1$

$$\begin{aligned}
&X=\left(\frac{3 \lambda+4}{\lambda+1}, \frac{5 \lambda+7}{\lambda+1}, \frac{3 \lambda+1}{\lambda+1}\right)\
&\text { Now } P X \perp A B \Rightarrow \overrightarrow{P X} \cdot \overrightarrow{A B}=0\
&\begin{aligned}
&\left(\frac{3 \lambda+4}{\lambda+1}-1\right) \cdot(4-3)+\left(\frac{5 \lambda+7}{\lambda+1}-0\right)(7-5) \
&+\left(\frac{3 \lambda+1}{\lambda+1}-3\right) \cdot(1-3)=0
\end{aligned}
\end{aligned}$$

$$\begin{aligned}
&\begin{aligned}
& \frac{2 \lambda+3}{\lambda+1}+\frac{10 \lambda+14}{\lambda+1}+\frac{4}{\lambda+1}=0 \
& \Rightarrow \frac{12 \lambda+21}{\lambda+1}=0 \Rightarrow \lambda=\frac{-7}{4} \
& X=\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)
\end{aligned}\
\&X \text { is mid point of } P Q\
\&\begin{aligned}
& Q \equiv\left(2 \cdot \frac{5}{3}-1,2 \cdot \frac{7}{3}-0,2 \cdot \frac{17}{3}-3\right) \equiv(\alpha, \beta, \gamma) \
& \Rightarrow \quad \alpha+\beta+\gamma=\frac{2(5+7+17)}{3}-4=\frac{58}{3}-4=\frac{46}{3}
\end{aligned}
\end{aligned}$$

Solutions · 1