Find the equation of the plane containing the point A115 and

Solution

First, let us introduce a couple of vectors. Let a=(1,-1,5). This is the direction vector of the line. Leb b=(2, -4, 2). This is the normal vector of the plane 2x-4y+2z=9. Now, let c be a normal vector of the plane we\'re looking for. It is easy to see that two planes are perpendicular if and only if their normal vectors are perpendicular. Therefore, c and b are perpendicular. Also, since the plane we\'re looking for contains the given line, this line is perpendicular to the normal of this plane. In other words, c and a are perpenducilar. So we want to find a vector c which is perpendicular to both a and b. One of such vectors is their vector product, so we can just find it: | i j k | | 3 2 -1| = i*(2*2-1*4) -j*(3*2+1*2) + k*(-3*4-2*2) = -8j - 16k | 2 -4 2| To make things more simple, let us divide this vector by -8 (it won\'t change its direction, and direction is the only thing important in a normal vector). So we get c=(0, 1, 2). This is the normal to the plane we are looking for. Also, this plain contains point A(1,-1,5), because the line contains it, and the plane contains the line. So we get the equation: 0*(x-(-1)) + 1*(y-5) + 2*(z-2) = 0 y + 2z - 9 = 0