Find the equation of the plane which is perpendicular to the vector 𝐀 equals five 𝐢 minus seven 𝐣 minus three 𝐤 and passes through the point 𝐵 negative five, five, nine.
Okay, here we’re trying to find the equation of a plane. And let’s say that this is that plane. And we’re told that this vector 𝐀 is perpendicular to our plane. That means if we sketched in vector 𝐀, it would look something like this. We also know that our plane passes through the point called point 𝐵. What we have then is a vector that is normal or perpendicular to our plane and a point the plane passes through. This is actually all we need to determine the equation of our plane.
Let’s recall the vector form of a plane’s equation. This form tells us that the dot product of a vector normal to the plane and a vector to a general point in it is equal to the dot product of that normal vector and a vector to a known point in the plane. In our case, the vector 𝐀 is our normal vector. And if we were to set up a coordinate frame, say, here, then a vector from the origin of that frame to point 𝐵 we could call 𝐫 zero. That’s our vector to a known point in the plane. 𝐫 zero then has components that are equal to the coordinates of point 𝐵. And since our normal vector 𝐧 is identical to the vector 𝐀, it has components five, negative seven, negative three.
We can now use these values in our vector form to solve for the equation of our plane. The normal vector dotted with a vector to a general point in our plane looks like this and that normal vector dotted with a vector to point 𝐵 looks like this. Carrying out these dot products, we get five 𝑥 minus seven 𝑦 minus three 𝑧 equals negative 25 minus 35 minus 27, which adds up to negative 87. If we now add positive 87 to both sides of this equation, then we find that five 𝑥 minus seven 𝑦 minus three 𝑧 plus 87 equals zero. This is what is called the general form of the equation of our plane.