Question

The equation of line $AB$ is $\frac{x}{2}=\frac{y}{-3}=\frac{z}{6}$. Through a point $P(1,2,5)$, line $PN$ is drawn perpendicular to $AB$ and line $PQ$ is drawn parallel to the plane $3x+4x+5z=0$ to meet $AB$ at $Q$. Then

• A. Coordinates of $N$ are $(\frac{52}{49},-\frac{78}{49},\frac{156}{49})$
• B. Equation of line $NQ$ is $3(x-3)=-(2y+9)=z-9$
• C. Equation of line $NQ$ is $3(x-3)=(2y+9)=z-9$
• D. Coordinates of $Q$ are $(3,\frac{9}{2},9)$

Answer 1

Answer: (A), (B)

The correct options are
A Coordinates of $N$ are $(\frac{52}{49},-\frac{78}{49},\frac{156}{49})$
B Equation of line $NQ$ is $3(x-3)=-(2y+9)=z-9$

Since, $N$ lies on line $AB$
and direction ratios of line $AB$ is $<2,-3,6>$
So, coordinates of $N\equiv (2r,-3r,6r)$
(where $r\in R$ be any constant)
$\therefore$ d.r.'s of $PN$ is $<2r-1,-3r-2,6r-5>$
Since, $PN\perp AB$
$\therefore 2(2r-1)-3(-3r-2)+6(6r-5)=0\Rightarrow r=\frac{26}{49}$
So, $N\equiv (\frac{52}{49},-\frac{78}{49},\frac{156}{49})$

Now, let coordinates of $Q$ be $(2k,-3k,6k)$
(where $k\in R$ be any constant)
and d.r.'s of perpendicular of the given plane is $<3,4,5>$
Then d.r.'s of $PQ$ is $<2k-1,-3k-2,6k-5>$
Since, $PQ\parallel$ plane
$\therefore 3(2k-1)+4(-3r-2)+5(6k-5)=0\Rightarrow k=\frac{3}{2}$
So, $Q\equiv (3,-\frac{9}{2},9)$
$\therefore$ Equation of $NQ$ is $\frac{x-3}{95}=\frac{y+9/2}{-285/2}=\frac{z-9}{285}$