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Find the equation of the plane …

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Find the equation of the plane Through the point (4,-3,2) And perpendicular to the line of intersection Of the plane x-y+2z-3=0 and 2x-Y-3z=0 Find the point of interaction of line x-1/1=y-2/3=z+1/-9 And the plane obtained above

Posted by Disha Yadav 7 years, 8 months ago

CBSE > Class 12 > Mathematics

1 answers

Kriti Sagar 7 years, 8 months ago

In this case ,eqn of plane will be (x-y+2z-3)+®(2x-y-3z)=0 U can find value of ® by putting value (4,-3,2) in place of (x,y,z) respectively in the eqn nd after that put value of ® in eqn. That will be the eqn of plane

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