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  • Help me please, - Three dimensional Coordinate Geometry - BITSAT

The line \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5} is parallel to the plane:

  • Option 1)

    3x+4y+5z=7

  • Option 2)

    2x+y-2z=0

  • Option 3)

    x+y-z=2

  • Option 4)

    2x+3y+4z=0

 

Answers (1)

best_answer

As we learnt in

Angle between line and Plane (Cartesian form ) -

The angle between a line and 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 is given by

\sin \Theta = \frac{a_{1}a+b_{1}b+c_{1}c}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a^{2}+b^{2}+c^{2}}}

-

 

 here a=3, b=4, c=5 & \theta=0

So, 3a_{1}+4b_{1}+5c_{1}=0

Only (2,1,-2) among the given options satify it 


Option 1)

3x+4y+5z=7

Option is incorrect

Option 2)

2x+y-2z=0

Option is correct

Option 3)

x+y-z=2

Option is incorrect

Option 4)

2x+3y+4z=0

Option is incorrect

Posted by

Aadil

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