Question

A line is drawn from the point P(1,1,1) and perpendicular to a line with direction ratios (1,1,1) to interact the plane x + 2y + 3z = 4 at Q. The locus of point Q is

• A. $x=\frac{y-5}{-2}=z+2$
• B. $\frac{x}{-2}=y-5=z+2$
• C. x = y = z
• D. $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$

Answer 1

Answer: (A)

$x=\frac{y-5}{-2}=z+2$

Given,

a Line $L$ , say , is drawn from a point $P\equiv (1,1,1)$ such that it is,

a) perpendicular to another line $L^{{}^{\prime }}$ having direction ratios $(1,1,1)$,

b) and ,also intersects a plane $P:=x+2y+3z=4$ at a point $Q$ .

And we are asked to find the locus of such a $Q$.

Say the direction ratio of the line $L$ is $(\alpha ,\beta ,\gamma )$ and since it is normal to line $L^{{}^{\prime }}$ with direction ratio $(1,1,1)$ by condition of perpendicularity

$\alpha .1+\beta .1+\gamma .1=0$

$\alpha +\beta +\gamma =0-\left(i\right)$

The line equation of $L$ is

$\frac{x-1}{\alpha }=\frac{y-1}{\beta }=\frac{z-1}{\gamma }=k$ (say)

$Q$ lies on this line , hence its coordinates will be of the form

$x=\alpha k+1-\left(ii\right)y=\beta k+1-\left(iii\right)z=\gamma k+1-\left(iv\right)$

Adding $\left(ii\right),\left(iii\right),\left(iv\right)$ we have

$x+y+z=k(\alpha +\beta +\gamma )+3$

$\Rightarrow x+y+z=3(\because \alpha +\beta +\gamma =0$ from $\left(i\right))-(v)$

Thus $Q$ will be on the plane given by equation $\left(v\right)$ & let's it as $P_1$ .However we are already given that $Q$ lies on the plane $P_0$.So how can be $Q$ on both the planes?

It can be on both the planes if the planes are intersecting . So basically $Q$ lies on the line of intersection of two planes $P_0$ & $P_1$ . Now the direction ratio of this line of intersection of two planes can be assumed to be $(l,m,n)$. And it will be perpendicular to the direction ratios of both the planes.Hence using the perpendicularity condition,

$l.1+m.1+n.1=0-\left(vi\right)l.1+m.2+n.3=0-\left(vii\right)\Rightarrow \frac{l}{1}=\frac{m}{-2}=\frac{n}{1}-\left(viii\right)$

Hence the direction ratios of the line formed by the intersection of the plane $P_0\&P_1$ is $(1,-2,1)$.

Now we need a point to write down the equation of this line & hence obtain the locus of $Q$.

We can set $x=0$, and putting it in the plane equation $P_0\&P_1$ we have,

$y+z=3-\left(ix\right)2y+3x=4-\left(x\right)$

Solving we obtain,

$y=5z=-2$

$\therefore$ the line equation will be

$\frac{x}{1}=\frac{y-5}{-2}=\frac{z+2}{1}$