Find the equation of the plane that contains the line L: (x,y,z) = (5, 1, 2) + t (3, -2, 4) , and the point (1, 2, 3) ?

When looking for the equation of a plane, the first thing we usually need to find is the normal vector. This determines the direction the plane "faces".

The normal vector will be orthogonal to the plane, and so will be perpendicular to any line or vector contained by the plane.

We are given one vector contained by the plane: <3, -2, 4>. This is the direction vector of the line contained by the plane, and thus this vector is contained by the plane.

A second vector can be found by subtracting one point from another, provided both are in the plane. We already know the point (1, 2, 3) is in the plane. We also know the point (5, 1, 2) is in the plane since this is the initial condition of the line for t=0.

So our second vector is found by subtracting (1, 2, 3) from (5, 1, 2):

<5-1, 1-2, 2-3> = <4, -1, -1>.

So our two vectors contained by the plane are <3, -2, 4> and <4, -1, -1>.

To find the normal vector of the plane, we take the cross product of two vectors contained by the plane.

<3, -2, 4> X <4, -1, -1> = <6, 19, 5>

so the plane's equation will be 6x + 19y + 5z = C where C is a constant.

To find C, just plug in one of the points on the plane ( (1, 2, 3) should work ).

6(1) + 19(2) + 5(3) = C

6 + 38 + 15 = C

C = 59

so the equation will be 6x + 19y + 5z = 59

Hope this helps.