Question

The line passing through the points $(5,1,a)$ and $(3,b,1)$ crosses the yz-plane at the spot $(0,\frac{17}{2},\frac{-13}{2})$. Then,

• A. $a=2,b=8$
• B. $a=4,b=6$
• C. $a=6,b=4$
• D. $a=8,b=2$

Answer 1

The correct option is C $a=6,b=4$

Equation of line passing through $(5,1,a)$ and $(3,b,1)$

is given by

$\frac{x-5}{3-5}=\frac{y-1}{b-1}=\frac{z-a}{1-a}$

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

in $yz$ plane $x=0$, And, $\frac{z-a}{1-a}=\frac{5}{2}$

$\frac{0-5}{-2}=\frac{y-1}{b-1}$

$\frac{5}{2}(b-1)+1=y$ $z=a+\frac{5}{2}(1-a)$__(II)

$\therefore y=\frac{5}{2}(b-1)+1$__(I)

But given $P(0,\frac{17}{2},\frac{-13}{2})$

$\frac{17}{2}=\frac{5}{2}(b-1)+1$ , $\frac{-13}{2}\frac{-5}{2}=a-\frac{5a}{2}$

$\Rightarrow \frac{15}{2}=\frac{5}{2}(b-1)$ $\Rightarrow -9=\frac{-3a}{2}$

$\therefore b=3+1$ $\Rightarrow a=6$

$\therefore b=4$

Hence, $a=6,b=4$