Question Video: Finding the Equation of a Plane Given the Normal Vector and a Point on the Plane Mathematics • Third Year of Secondary School

Video Transcript

Give the general form of the equation of the plane with normal vector 10, eight, three that contains the point 10, five, five.

Let’s begin by assuming this is our three-dimensional plane as shown. We are told that it has a normal vector, that is, a vector perpendicular to the plane, that has components 10, eight, three. The plane also contains a point that we will call 𝑃 sub zero with coordinates 10, five, five. And we are asked to find the general form of the equation of this plane. This is written 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 minus 𝑑 equals zero, where the vector normal to the plane is equal to 𝑎, 𝑏, 𝑐 and 𝑑 is equal to the scalar or dot product of the normal vector 𝐧 and the vector 𝐫 sub zero. In this question, we are not given a vector 𝐫 sub zero but, as previously stated, are given a point 𝑃 sub zero.

Drawing a vector from the origin of our coordinate plane to the point 𝑃 sub zero, then this vector 𝐫 sub zero will have components equal to the coordinates of 𝑃 sub zero. We are now in a position to calculate the value of 𝑑 by finding the dot product. 𝑑 is equal to the dot product of vectors 10, eight, three and 10, five, five. To calculate this, we find the product of the corresponding components and then the sum of these three values. This is equal to 10 multiplied by 10 plus eight multiplied by five plus three multiplied by five. This simplifies to 100 plus 40 plus 15, which is equal to 155.

We now have values of 𝑎, 𝑏, 𝑐, and 𝑑 which we can substitute into the general form. The equation of the plane with normal vector 10, eight, three that contains the point 10, five, five is 10𝑥 plus eight 𝑦 plus three 𝑧 minus 155 is equal to zero.