Question

Determine a scalar equation for the plane that passes through the point (2, 0, −1) and is perpendicular to the line of intersection of the planes
2x + y – z + 5 = 0 and x + y + 2z + 7 = 0.

Answer 1

take the cross product of the 2 planes?

Answer 2

which would give you Ax + By + Cz + D=0
then sub in the point (2, 0, −1) into the equation to find D

Answer 3

need help in finding cross product?

Answer 4

I already did that I just didnt know how to continue
Thank you very much :)

Answer 5

so what did you get for the vector after the cross product?

Answer 6

well, once you have that vector from the cross product, that will be the normal vector for you plane
you can get the scalar form of the plane from using the fact that dotting the normal vector with a vector within its own plane is zero

Answer 7

\[\vec n\cdot(\vec P-\vec P_0)=0\]where \(\vec P=\langle x,y,z\rangle\) and \(\vec P_0=\langle 2,0,-1\rangle\)

Answer 8

I turned the equations into 2x +y -z=-12 and x +y +2z=-7
Then I crossed (2,1,-1) with (1,1,2) and got (-3,5,-1)

Answer 9

where did the -12 come from?

oh well, makes no difference... yes I agree that is the cross product, which is the normal vector of your new plane \(\vec n=\langle-3,5,-1\rangle\)

Answer 10

Oh woops typo lol
I was looking at another question I did previously >.<

Answer 11

Now pick a random point in the plane. Call its position vector \(\vec P=\langle x,y,z\rangle\)
let the position vector of the given point be \(\vec P_0=\langle 2,0,-1\rangle\)

notice that the difference of these two position vectors\[\vec P-\vec P_0\] will be a vector that lies in the plane

if the vector lies in the plane, then it is perpendicular to the normal vector we just found

if two vectors are perpendicular, their dot-product is zero, hence\[\vec n\cdot(\vec P-\vec P_0)=0\]using this formula will give you the scalar equation of the plane

Answer 12

Alright, so I did (-3,5,-1) . (X-2,y, z +1)
Which ended up giving me -3x +5y -z +5=0