Hi i have a plane : $P1 x-2y+z=4$ and the points $A(i+j+k)$ , $B (2i -3j-k)$ and $C (5i-j)$

My first part of the question is to find $P2$ (passes through A and B and is perpendicular to P1), which i found by finding the displacement of between A and B. and then finding the cross product of this displacement and the normal of $P1$

My final result was $P2$: $-8x -3y +2z=-9$, is this correct?

Then I had to find the equation of $P3$ which is perpendicular to both planes and passes through point C, to do this i first found the cross product of the two plane normals and used this to find the value that any point on the plane would produce when multiplied with this cross product. Resulting in $-x -10y-19z=5$

is this the correct result? im asking because vectors are a huge weakpoint of mine and while i found other questions relating to this i found the answers too complicated to understand


1 Answer 1


You have correctly computed the second plane $P_2$. Now for $P_3$, we can simply find the normal vector $\vec{n_3}$ as cross product of normal vectors of $P_1$ and $P_2$:

$$\vec{n_3} = \vec{n_1} \times \vec{n_2}$$

After calculation, you get $\vec{n_3}=\langle 1,10,19\rangle$ Now the equation of plane 3 is $x+10y+19 =d$. Since $C$ lies on this plane, it satisfies this. We get $d = -5.$

This also you have found correctly!


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