Question

In the projection of point $P(\mathop p\limits^ \to )$on the plane$\mathop r\limits^ \to .\mathop n\limits^ \to = q$ is the points $S(\mathop s\limits^ \to )$, then

Answer 1

The projection of a point is its shadow on the plane or central projection.
If C is a point, called the centre of projection then the projection of a point P different from C onto a plane that does not contain C is the interaction of the line CP with the plane.

Complete step-by- step solution:

Let us draw a plane and the projection of point $P(\mathop p\limits^ \to )$on the plane is $\mathop s\limits^ \to$.
The intersection is $\mathop r\limits^ \to .\mathop n\limits^ \to = q$
As the line is normal to the plane i.e. perpendicular to the plane and vector $\mathop P\limits^ \to$ is passing through the plane and parallel to $\mathop n\limits^ \to$
$E{q^n}$ of such a line is$\overrightarrow r = \overrightarrow p + \lambda \overrightarrow n ……...(1)$
Given, $\mathop r\limits^ \to .\mathop n\limits^ \to = q........(2)$
As the line is passing through the plane, then the equation (1) will be satisfying equation (2) and that point $\mathop r\limits^ \to = \mathop s\limits^ \to$
Substituting equation (1) in (2), we get:
$\Rightarrow$ $(\mathop p\limits^ \to + \lambda \mathop n\limits^ \to )\mathop n\limits^ \to = q$
To find the value of $\lambda$, simplify the above term then we get it as
$\Rightarrow \mathop p\limits^ \to . \mathop n\limits^ \to + \lambda \mathop n\limits^ \to .\mathop n\limits^ \to = q$
As $\left[ {\overrightarrow n .\overrightarrow n = {{\left| {\mathop n\limits^ \to } \right|}^2}} \right]$ , we get:
$\Rightarrow \lambda {\left| {\mathop n\limits^ \to } \right|^2} = q - \mathop p\limits^ \to .\mathop n\limits^ \to$
$\Rightarrow \lambda = \dfrac{{q - \mathop p\limits^ \to .\mathop n\limits^ \to }}{{{{\left| {\mathop n\limits^ \to } \right|}^2}}}$______ (3) {On RHS ${\left| {\mathop n\limits^ \to } \right|^2}$ will be in division as it was multiplication on LHS}
Now using equation (3) in (1), we get:
$\mathop r\limits^ \to = \mathop p\limits^ \to + (\dfrac{{q - \mathop p\limits^ \to .\mathop n\limits^ \to }}{{{{\left| {\mathop n\limits^ \to } \right|}^2}}})\mathop n\limits^ \to$
We know that$\overrightarrow r = \overrightarrow s$, hence:
$\mathop s\limits^ \to = \mathop p\limits^ \to + (\dfrac{{q - \mathop p\limits^ \to .\mathop n\limits^ \to }}{{{{\left| {\mathop n\limits^ \to } \right|}^2}}})\mathop n\limits^ \to$

Note: Two planes are parallel if they have the same normal vector (i.e. their normal vectors are parallel). If two planes are not parallel, then they intersect in a line.If any line passes through a plane then it always satisfies the equation of that plane.