Find a standard form equation of a plane which is perpendicular to the line (x, y, z) = (1, 2, 1) + t(3, 2, -1) such that the plane also passes through the point (5, -1, 3).
Solution: The vector (3, 2, -1) is parallel to the given line and therefore is perpendicular to any line which is perpendicular to this this line.Therefore n = (3, 2, -1) is a normal for the line whose standard form equation we are asked to write. The equation has the form n1x + n2y + n3z = n • p where P is the point known to be on the line. For the normal we have found and the point P (5, -1, 3) we get
n
- p=(3,2,-1) (5,-1,3)=3(5)+2(-1)+(-1)(3)=15-2-3=10
HOW DOES THE VECTOR (3,2,-1) WHICH IS PARALLEL TO THE GIVEN LINE, BECOME PERPENDICULAR (COMPONENTS OF NORMAL) ?