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Equation of the plane which passes through the point of intersection of lines

$\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}\:$ and $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}\:$ and has the largest distance from the origin is :

Option 1)
$7x+2y+4z=54$

Option 2)
$3x+4y+5z=49$

Option 3)
$4x+3y+5z=50$

Option 4)
$5x+4y+3z=57$

Answers (1)

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

Point of interactionof two lines

$\frac{x-1}{3}= \frac{y-2}{1}=\frac{z-3}{2} =k$

$x= 3k+1,\, \, \, y = k+2,\, \, \, z=2k+3$

and

$\frac{x-3}{1}= \frac{y-1}{2}=\frac{z-2}{3} =l$

$x=l+3; y=2l+1; z=3l+2$

$3k+1=l+3;k+2 = 2l+1$

$3k-l=2\ and\ k-2l = -1$

$6k-2l =4$

$k-2l=-1$

$k=1, l=1$

Point is (4,3,5)

Plane $4x+3y+5z=50$ has maximum distance from origin

Option 1)

$7x+2y+4z=54$

This option is incorrect

Option 2)

$3x+4y+5z=49$

This option is incorrect

Option 3)

$4x+3y+5z=50$

This option is correct

Option 4)

$5x+4y+3z=57$

This option is incorrect

Posted by

Sabhrant Ambastha

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