Question

The image of the point with position vector $[ \vec{i}+3\vec{k} ]$ in the plane $[\vec{r}.(\vec{i}+\vec{j}+\vec{k})=1 ]$ is-
A. $\vec{i}+2\vec{j}+\vec{k}$
B. $ \vec{i}-2\vec{j}+\vec{k}$
C. $-\vec{i}-2\vec{j}+\vec{k}$
D. $\vec{i}+2\vec{j}-\vec{k}$

Answer 1

Hint: The object having both magnitude and direction are called vectors. The position vector is that vector that locates the position of any point on any plane from any origin. The direction of the position vector is from the origin towards the point. The point can also be a moving body.

Complete step by step solution: Let on the plane $[\vec{r}.(\vec{i}+\vec{j}+\vec{k})=1 ]$ the image of point P$ [ (\vec{i}+3\vec{k}) ]$ is Q.










Hence, we can see that the PQ is normal/perpendicular to the plane$ [ \vec{r}.(\vec{i}+\vec{j}+\vec{k})=1 ]$. Since PQ is passing through P and is perpendicular to the given plane, the coordinates of P is (1,0,3), so the equation of PQ will be –
r=$(\vec{i}+3\vec{k})+\lambda(\vec{i}+\vec{j}+\vec{k})$
Since Q also lies on the line PQ, so the position vector of Q is$ [ (\vec{i}+3\vec{k})+\lambda(\vec{i}+\vec{j}+\vec{k})]$
⇒ $ (1+\lambda)\vec{i}+\lambda\vec{j}+(3+\lambda)\vec{k}$
As we know that the R is the midpoint of PQ, therefore position vector of R is
⇒ $\dfrac{(1+\lambda)\vec{i}+\lambda\vec{j}+(3+\lambda)\vec{k}+\vec{i}+3\vec{k}}{2}$
Or, $ (\dfrac{\lambda+2}{2})\vec{i}+(\dfrac{\lambda}{2})\vec{j}+(3+\dfrac{\lambda}{2})\ \vec{k}$
Or, $(\dfrac{\lambda}{2}+1)\vec{i}+(\dfrac{\lambda}{2})\vec{j}+(3+\dfrac{\lambda}{2})\ \vec{k}$
Since, R lies on plane$ [ \vec{r}.(\vec{i}+\vec{j}+\vec{k})=1]$
Hence,
⇒ $\left[(\dfrac{\lambda}{2}+1)\vec{i}+(\dfrac{\lambda}{2})\vec{j}+(3+\dfrac{\lambda}{2})\ \vec{k}\right].(\vec{i}+\vec{j}+\vec{k})=1$
⇒ $\left[(\dfrac{\lambda}{2}+1+\dfrac{\lambda}{2}+3+\dfrac{\lambda}{2})\ \right]=1$
⇒ λ = ―2
So, the position vector of Q is-
$(\vec{i}+3\vec{k})-2(\vec{i}+\vec{j}+\vec{k})=\ -\vec{i}-2\vec{j}+\vec{k} $
The image Q of the point with position vector$ [(\vec{i}+3\vec{k}) ]$ in the plane $[ \vec{r}.(\vec{i}+\vec{j}+\vec{k})=1 ] is [ -\vec{i}-2\vec{j}+\vec{k}]$

Thus, Option (C) is correct.

Note: When there is a change in the position of the position vector the difference in the change in the position of position vector is called displacement vector. The direction of the displacement vector is from the initial towards the final position. There are different types of vectors like unit vector, zero vector, collinear vector, equal vector, etc.