Question

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

Answer 1

Let the image of a point having position vector $3\hat{i}-2\hat{j}+\hat{k}$ in the plane is $p^{\prime },q^{\prime },r^{\prime }$ and $p,q,r$ is the midpoint of the line joining the image of the point and the point hich is present on the plane

The vector $(3-p)\hat{i}-(2-q)\hat{j}+(1-r)\hat{k}$, must be parallel to the normal vector of plane that is $3\hat{i}-\hat{j}+4k$

$(3-p)\hat{i}-(2-q)\hat{j}+(1-r)\hat{k}=$$x(3\hat{i}-\hat{j}+4k)\dots \left(1\right)$,

Where $x$ is a constant.

and also $p\hat{i}+q\hat{j}+r\hat{k}$ is in the plane

therefore, $(p\hat{i}+q\hat{j}+r\hat{k})\cdot$$(3\hat{i}-\hat{j}+4k)=2$

$3p-q+4r=2$

From (1),

$3(3-3x)-(2-x)+4(1-4x)=2⟹x=\frac{11}{24}$

$p=\frac{13}{8},q=\frac{37}{24},r=\frac{-5}{6}$

we know,

$p=\frac{3+p^{\prime }}{2}$

$q=\frac{-2+q^{\prime }}{2}$

$r=\frac{1+r^{\prime }}{2}$

$p^{\prime }=\frac{1}{4},q^{\prime }=\frac{61}{12},r^{\prime }=\frac{-8}{3}$

therefore the position vector of image is

$\frac{1}{4}\hat{i}+\frac{61}{12}\hat{j}-\frac{8}{3}\hat{k}$