Question

Find the image of the point having position vector $\hat{i}+3\hat{j}+4\hat{k}$ in the plane $\overrightarrow{r}.(2\hat{i}-\hat{j}+\hat{k})+3=0$.

Answer 1

Let B be the point which is the image of the given point A=i+3j+4k.

So, $AB$ is normal to the given plane $AB$

So, equation of AB is $r=(i+3j+4k)+\lambda (2i-j+k)$

Now, Let us consider the $B=(1+2\lambda )i+(3-\lambda )j+(4+\lambda )k$

We know that the midpoint of AB lies on the plane whose equation is given

Midpoint $T=(A+B)/2\Rightarrow T=(\lambda +1)i+(3-\frac{\lambda }{2})j+(4+\frac{\lambda }{2})k$

Put T in the equation of the plane, we will get the value of $\lambda$ and using that value of $\lambda$ we get B.

$r.(2i-j+k)+3=0$

$(\lambda +1)i+(3-\frac{\lambda }{2})j+(4+\frac{\lambda }{2})k.(2i-j+k)+3=0$

On solving

$2(\lambda +1)-(3-\frac{\lambda }{2})+(4+\frac{\lambda }{2})+3=0$

$2\lambda +\frac{\lambda }{2}+\frac{\lambda }{2}=-6\lambda =-2$

So, $B=\left[\right(1-4)i+(3-(-2)j+(4+(-2\left)\right)k]=-3i+5j+2k$

Hence the image of the point having position vector $i+3j+4k$ in the plane$r.(2i-j+k)+3=0$ is

$B=-3i+5j+2k$