To find the
equation of plane P1, namely ax+by+cz=1:
2a + b
=1
a+b+c
= 1
a + 7b +3c
=1
which gives a=3/5, b=-1/5, c=3/5 and hence the normal to P1 is the vector (3, -1, 3).
This can also be found from the cross product
® AB
×
® AC
= (-1, 0, 1) ×(-1, 6, 3)
.
Suppose the plane P2 has equation lx+my+nz=1, then for the
planes P1 and P2 to be perpendicular the point V must
satisfy this equation. The normal vector to P2, i.e. (l,m,n)
must be perpendicular to (3,-1,3) so 3l-m+3n = 0. As A and B
are on P2 we have:
2l+m
=1
l+m+n
=1
3l-m+3n
=0
Hence l=1/8, m=3/4 and n=1/8 so the equation
of P2 is x +6y+z=8. The coordinates of V must satisfy this
equation giving an infinite set of points on the plane P2.