The distance of the point (1, 0, 2) from the line point of intersection of the line

\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}      and the plane  x-y+z=16,  is:

  • Option 1)

    2\sqrt{14}

  • Option 2)

    8

  • Option 3)

    3\sqrt{21}

  • Option 4)

    13

 

Answers (1)

As we learnt in 

Intersection of line and plane -

Let the line

\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} plane

a_{1}x+b_{1}y+c_{1}z+d=0 intersect at P

to find P assume general point on line as \left ( x_{1}+\lambda a_{1}y_{1} +\lambda b_{1}z_{1}+\lambda c_{1}\right )

now put it in plane to find \lambda,

a_{1}\left ( x_{1}+\lambda a \right )+b_{1}\left ( y_{1}+\lambda b \right )+c_{1}\left ( z_{1}+\lambda c \right )+d=0

-

 

 Point of intesection of line

\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=k

x=3k+2, y=4k-1,z=12k+2

x-y+z=16

3k+2-4k+1+12k+2=16

11k-11\Rightarrow    k=1

Point is (5,3,14)

Distance between (5,3,14) and (1,0,2)

is \sqrt{4^{2}+3^{2}+12^{2}}=13


Option 1)

2\sqrt{14}

This is incorrect option

Option 2)

8

This is incorrect option

Option 3)

3\sqrt{21}

This is incorrect option

Option 4)

13

This is correct option

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