Question

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

• A. 15
• B. 1
• C. 2
• D. 16

Answer 1

Answer: (A), (B)

1

Equation of the line passing through the point $P(1,4,4)$ is $\frac{x-1}{a}=\frac{y-4}{b}=\frac{z-3}{c}....\left(i\right)$
If equation (i) is perpendicular to lines $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-2}{4}$
and $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z+1}{-2}$
$\therefore 2a+b+4c=0....\left(ii\right)$
and $3a+2b-2c=0.....\left(iii\right)$
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) and or , (11, -2, 2)
$\therefore a_1+a_2+a_3=-9+20+4$
or, $11-12+2$
= 15 or 1