Question

A line is drawn from the point $P(1,1,1)$ and perpendicular to a line with direction ratios $(1,1,1)$ to intersect 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)

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

Let the co-ordinate of $Q$ be $(h,k,l)$

Point $Q$ lies on the plane $x+2y+3z=4$

$\therefore h+2k+3l=4-\left(1\right)$

$\overrightarrow{PQ}=\overrightarrow{Q}-\overrightarrow{P}$

$=(h-1)\hat{i}+(k-1)\hat{j}+(l-1)\hat{k}$

Let $\overrightarrow{u}$ be vector with DR $(1,1,1)$

$\therefore \overrightarrow{u}=\hat{i}+\hat{j}+\hat{k}$

$\therefore \overrightarrow{PQ}$ is $\perp$ to $\overrightarrow{u}$

$\therefore \overrightarrow{PQ}.\overrightarrow{u}=0$

$(h-1).1+(k-1).1+(l-1).1=0$

$h+k+l=3-\left(2\right)$

locus = line of intersection of (1) & (2)

$\overrightarrow{n_1}$ (vector of (1) plane) $=\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{n_2}$ (vector of (2) plane) $=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{a}\perp \overrightarrow{n_1}$ and $\overrightarrow{a}\perp \overrightarrow{n_2}$

where $\overrightarrow{a}$ is vector of required locus

$\overrightarrow{a}=\overrightarrow{n_2}\times \overrightarrow{n_1}$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& 2& 3\end{array}\right|$

$=\hat{i}-2\hat{j}+\hat{k}(DR=1,-2,1)$

$h+2k+3l=4-\left(1\right)$

$h+k+l=3-\left(2\right)$

putting $h=0$ & solving for $k$ & $l$

$2k+3l=4-\left(3\right)$

$2k+2l=6-\left(4\right)$

From (3) & (4)

$l=-2,k=5$

$\frac{x-0}{1}=\frac{y-5}{-2}=\frac{z+2}{1}$