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

Option 1)
$a=8,b=2$

Option 2)
$a=2,b=8$

Option 3)
$a=4,b=6$

Option 4)
$a=6,b=4$

Answers (1)

best_answer

As we learnt in

Cartesian eqution of a line -

The equation of a line passing through two points $A\left ( x_{0},y_{0},z_{0} \right )$ and parallel to vector having direction ratios as $\left ( a,b,c \right )$ is given by

$\frac{x-x_{0}}{a}= \frac{y-y_{0}}{b}= \frac{z-z_{0}}{c}$

The equation of a line passing through two points $A\left ( x_{1},y_{1},z_{1} \right )\, and \, B\left ( x_{2},y_{2},z_{2} \right )$ is given by

$\frac{x-x_{1}}{x_{2}-x_{1}}= \frac{y-y}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$

- wherein

Equation of line is

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

If it crosses y-z plane, x=0

$2\lambda +5=0 \, \, \, = > \lambda =\frac{-5}{2}$

$y=\lambda \left ( 1-b \right )+1=\frac{17}{2}\, \Rightarrow b= 4$

$z=\lambda(a-1)+a=\frac{-13}{2}$

$\Rightarrow a = 6$

Option 1)

$a=8,b=2$

Incorrect Option

Option 2)

$a=2,b=8$

Incorrect Option

Option 3)

$a=4,b=6$

Incorrect Option

Option 4)

$a=6,b=4$

Correct Option

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