The distance of the point $(3,4,5)$ from the point of intersection of the line…
The distance of the point $(3,4,5)$ from the point of intersection of the line
$\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ and plane $x+y+z=2$ is
6 units
13 units
10 units
7 units
Solution
Given $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda \quad \ldots$ (say) ...(1)
Hence coordinates of any point on this line are $\therefore x=\lambda+3, y=2 \lambda+4, z=2 \lambda+5$
Since this point lies on the plane, we write
$\begin{array}{l}
(\lambda+3)+(2 \lambda+4)+(2 \lambda+5)=2 \\
5 \lambda+12=2 \Rightarrow \lambda=-2
\end{array}$
Hence coordinates of point of inter section are $\equiv(-2+3,-4+4,-4+5)$ i.e. $(1,0,1)$ $\therefore$ Distance between $(3,4,5)$ and $(1,0,1)$ is
$=\sqrt{(3-1)^{2}+(4-0)^{2}+(5-1)^{2}}=6$