### 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.

Letβs 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 π 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 ππ₯ plus ππ¦ plus ππ§ minus π equals zero, where the vector
normal to the plane is equal to π, π, π and π is equal to the scalar or dot
product of the normal vector π§ and the vector π« sub zero. In this question, we are not given a vector π« sub zero but, as previously stated,
are given a point π sub zero.

Drawing a vector from the origin of our coordinate plane to the point π sub zero,
then this vector π« sub zero will have components equal to the coordinates of π sub
zero. We are now in a position to calculate the value of π by finding the dot product. π 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 π, π, π, and π 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 10π₯ plus eight π¦ plus three π§ minus 155 is equal to zero.