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)
S Sabhrant Ambastha

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

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