Cartesian Equation of Plane

Find the cartesian equation of a plane which passes through the point (3,-4,1) and is parallel to the plane containing the point (1,2,-1) and the line** r**=t(1,1,1).

The answer is 3x-2y-z=16. I don't know how to find the normal vector of the plane. I know that the fixed point is (3,-4,1). Please help.

I'm really weak in vectors, is there any way to improve my mastery of this topic?(Worried)

Re: Cartesian Equation of Plane

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Originally Posted by

**SkyCapri** Find the cartesian equation of a plane which passes through the point (3,-4,1) and is parallel to the plane containing the point (1,2,-1) and the line** r**=t(1,1,1).

The answer is 3x-2y-z=16. I don't know how to find the normal vector of the plane. I know that the fixed point is (3,-4,1). Please help.

the required normal is $\displaystyle n=<1,2,-1>\times<1,1,1>$.

Re: Cartesian Equation of Plane

Taking t= 0, one point on that line, and so in the plane, is (0, 0, 0). Taking t= 1, another point in the plane is (1, 1, 1). <1, 1, 1> is the vector from (0, 0, 0) to (1, 1, 1). <1, 2, -1> is the vector from (0, 0, 0) to (1, 2, -1). Those are two vectors in the plane so their cross product is normal to the plane.