Question

In ${{R}^{3}}$ , let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes ${{P}_{1}}:x+2y-z+1=0$ and ${{P}_{2}}:2x-y+z-1=0$ . Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane ${{P}_{1}}$ . Find the equation M.

Answer 1

Hint : As the line is at a constant distance from the given two planes, we can say that the line is parallel to both the planes. We know that for a general plane ax+by+cz+d=0, $a\widehat{i}+b\widehat{j}+c\widehat{k}$ represents the vector normal to the given plane, so from the equation of the two planes, we can easily find two vectors which are perpendicular to the planes and hence the line L. Take their cross product to find a vector along the line L, hence its direction cosines. Also, we know that the line passes through origin so get the equation of line, followed by a general point on it and use the formula of foot perpendicular to plane one to get the locus.

Complete step by step solution :
It is given that all the points on L are at a constant distance from the two planes ${{P}_{1}}:x+2y-z+1=0$ and ${{P}_{2}}:2x-y+z-1=0$ , so the line is parallel to the planes. We know that for a general plane ax+by+cz+d=0, $a\widehat{i}+b\widehat{j}+c\widehat{k}$ represents the vector normal to the given plane. Therefore, normal vector to the plane ${{P}_{1}}:x+2y-z+1=0$ is $\widehat{i}+2\widehat{j}-\widehat{k}$ and the normal to the plane ${{P}_{2}}:2x-y+z-1=0$ is $2\widehat{i}-\widehat{j}+\widehat{k}$ .
Therefore, the vector along line L is the cross product of the two normal vectors.
$\left( 2\widehat{i}-\widehat{j}+\widehat{k} \right)\times \left( \widehat{i}+2\widehat{j}-\widehat{k} \right)$
For finding the cross product we need to form the determinant, in which the first row is $\widehat{i},\widehat{j}\text{ and }\widehat{k}$ , second row is the coefficients of $\widehat{i},\widehat{j}\text{ and }\widehat{k}$ in first vector and third row consists of the coefficient of $\widehat{i},\widehat{j}\text{ and }\widehat{k}$ in second vector.
$\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\\ 2 & -1 & 1 \\\ 1 & 2 & -1 \\\ \end{matrix} \right|=\widehat{i}\left( 1-2 \right)-\widehat{j}\left( -2-1 \right)+\widehat{k}\left( 4+1 \right)=-\widehat{i}+3\widehat{j}+5\widehat{k}$
Therefore, DRs of the line L is -1, 3, 5. Also, the line passes through the origin.
Now we know that the equation of line with DRs as l,m,n and passing through origin is given by:
$\dfrac{x}{l}=\dfrac{y}{m}=\dfrac{z}{n}$ .
Therefore, the equation of line L is:
$\dfrac{x}{-1}=\dfrac{y}{3}=\dfrac{z}{5}$
Now the general point lying on Line L is:
$\dfrac{x}{-1}=\dfrac{y}{3}=\dfrac{z}{5}=\lambda$
$x=-\lambda$
$z=5\lambda$
$y=3\lambda$
The general point comes out as: $\left( -\lambda ,3\lambda ,5\lambda \right)$ .
Now we know that the foot of perpendicular (g,h,k) of a point (p,q,r) to a general plane is given by:
$\dfrac{g-p}{a}=\dfrac{h-q}{b}=\dfrac{k-r}{c}=\dfrac{-\left( ap+bq+cr+d \right)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$
So, let the general point on L be $\left( -\lambda ,3\lambda ,5\lambda \right)$ , so the foot of perpendicular on the plane ${{P}_{1}}:x+2y-z+1=0$ be (x,y,z), which will be general point for M.
$\dfrac{x-\left( -\lambda \right)}{1}=\dfrac{y-3\lambda }{2}=\dfrac{z-5\lambda }{-1}=\dfrac{-\left( 1\times \left( -\lambda \right)+2\times \left( 3\lambda \right)-1\times \left( 5 \lambda \right)+1 \right)}{{{1}^{2}}+{{2}^{2}}+{{\left( -1 \right)}^{2}}}$
$\Rightarrow x+\lambda =\dfrac{y-3\lambda }{2}=\dfrac{z-5\lambda }{-1}=\dfrac{-1}{4}$

$x+\lambda =\dfrac{-1}{4}$
$\Rightarrow x+\dfrac{1}{4}=\lambda$
$\dfrac{y-3\lambda }{2}=\dfrac{-1}{4}$
$\Rightarrow \dfrac{y+\dfrac{1}{2}}{3}=\lambda$
$\dfrac{z-5\lambda }{-1}=\dfrac{-1}{4}$
$\Rightarrow \dfrac{z-\dfrac{1}{4}}{5}=\lambda$
Therefore, we can conclude that the equation M is:
$x+\dfrac{1}{4}=\dfrac{y+\dfrac{1}{2}}{3}=\dfrac{z-\dfrac{1}{4}}{5}$

Note : In questions related to lines and planes the key thing is to remember the important formulas, like the foot of perpendicular to the plane, image of a point etc. Also, the other key point was to interpret the statement that the distance of each point is constant, which means they are parallel.