Question

A line L passing through the point $P(1,4,3)$ is perpendicular to both the lines
$\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-2}{4}\text{and}\frac{x+2}{3}=\frac{y-4}{2}=\frac{z+1}{-2}$.
If the position vector of the point Q on L is $(a_1,a_2,a_3)$ such that $P{Q}^2=357,\text{then}(a_1+a_2+a_3)$ can be

• A. 15
• B. 16
• C. 2
• D. 1

Answer 1

Answer: (A), (C)

2

Equation of the line passing through the point
$P(1,4,3)\text{is}\frac{x-1}{a}=\frac{y-4}{b}=\frac{z-3}{c}$...(i)
If equation (i) is perpendicular to lines $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-2}{4}\text{and}\frac{x+2}{3}=\frac{y-4}{2}=\frac{z+1}{-2}$
$\therefore 2a+b+4c=0$ ...(ii)
and $3a+2b–2c=0$ ...(iii)
Solving (ii) and (iii),
$\frac{a}{-2-8}=\frac{-b}{-4-12}=\frac{c}{4-3}$
$\Rightarrow \frac{a}{-10}=\frac{b}{16}=\frac{c}{1}=\lambda$ , say
$\therefore a=-10\lambda ,b=16\lambda ,c=\lambda$
Hence the equation of the line is
$\frac{x-1}{-10\lambda }=\frac{y-4}{16}=\frac{z-3}{\lambda }$
$\Rightarrow \frac{x-1}{-10}=\frac{y-4}{16}=\frac{z-3}{1}=r$, say ...(iv)
$\therefore$Any point Q on line (ii) is
$(1–10r,16r+4,r+3)$
Distance of Q from $P(1,4,3)=(1–10r–1{)}^2+(16r+4–4{)}^2+(r+3–3{)}^2$
$\Rightarrow r^2(100+256+1)=357$
$\Rightarrow r^2=1$
$\Rightarrow r=\pm 1$
$\therefore$The point Q is $(–9,20,4)\text{and or},(11,–12,2)$
$\therefore a_1+a_2+a_3=–9+20+4$
or, $11–12+2$
$=15\text{or}1$