Question

Find an equation for the plane which passes through the point p(1,3,3) and contains the line: l:x(t)=3t,y(t)=2t,z(t)=4+3t

Answer 1

Answer:

Step-by-step explanation:

Given that the plane passes through (1,3,3)

Also the plane contains the line x(t)=3t,y(t)=2t,z(t)=4+3t.

This means all points lying in this line will also lie in the plane.

Find out two points on this line.

First point: Let t =0.  Point is (0,0,4)

Next point : Let t = 1: Point is (3,2,7)

Now we have 3 non collinear points (0,0,4) (3,2,7) and (1,3,3) lying on the plane.

Equation of the plane is

$\left[\begin{array}{ccc}x-0&y-0&z-4\\3&2&3\\1&3&-1\end{array}\right] =0$

Simplify to get

x(-2-9)-y(-3-3)+(z-4)(9-2)=0

i.e -11x+6y+7z-28 =0

11x-6y-7z+28 =0 is the equation of the plane.