Inbrunstlr

2022-10-03

A plane passes through the point (1,1,1) and is perpendicular to each of the planes

3x−2y+3z+6=0 and 6x−2y−3z−6=0. Find its equation. The problem is I don't have an idea of the concept. All I know is that the normal of first equation is (3,−2,3) and that of the second is (6,−2,−3).

3x−2y+3z+6=0 and 6x−2y−3z−6=0. Find its equation. The problem is I don't have an idea of the concept. All I know is that the normal of first equation is (3,−2,3) and that of the second is (6,−2,−3).

Dayana Powers

Beginner2022-10-04Added 6 answers

Here’s a hint:

the normal of the third plane is perpendicular to both normals of the two given planes.

Use the cross product.

the normal of the third plane is perpendicular to both normals of the two given planes.

Use the cross product.

kasibug1v

Beginner2022-10-05Added 4 answers

So if T(x,y,z) is in this plane and A(1,1,1) then

$\overrightarrow{AT}=(x-1,y-1,z-1)=m(3,-2,3)+n(6,-2,-3)$

for some scalars m,n. Eliminate the scalars and you are done.

$\overrightarrow{AT}=(x-1,y-1,z-1)=m(3,-2,3)+n(6,-2,-3)$

for some scalars m,n. Eliminate the scalars and you are done.

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