Question
The equation of the plane passing through the points (1,-3,-2) and perpendicular to planes x +2y+2z=5 and 3x+3y+2z=8, is
Answer: Option A
:
A
l+2m+2n=0, 3l+3m+2n=0, l2+m2+n2=1, we get l, m, nfrom these equations and then putting the values in l(x-1)+m(y+3) +n(z+2)=0, we get the required result.
Trick: Checking conversely,
2(x)-4(y)+3(z)-8 =0,
So, it passes through given point.
1(2)+2(-4)+2(3) =0,
So, it is perpendicular to x+2y+2z=5.
3(2)+3(-4)+2(3)=0,
So, it is perpendicular to 3x+3y +2z=8.
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:
A
l+2m+2n=0, 3l+3m+2n=0, l2+m2+n2=1, we get l, m, nfrom these equations and then putting the values in l(x-1)+m(y+3) +n(z+2)=0, we get the required result.
Trick: Checking conversely,
2(x)-4(y)+3(z)-8 =0,
So, it passes through given point.
1(2)+2(-4)+2(3) =0,
So, it is perpendicular to x+2y+2z=5.
3(2)+3(-4)+2(3)=0,
So, it is perpendicular to 3x+3y +2z=8.
Was this answer helpful ?
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