Video Transcript
Find the general equation of the plane which passes through the point five, one, negative one and is parallel to the two vectors nine, seven, negative eight and negative two, two, negative one.
In this question, we’re asked to find the general equation of a plane. And to do this, we’re given that it passes through the point with coordinates five, one, negative one. And we’re given two vectors parallel to the plane, the vector nine, seven, negative eight and the vector negative two, two, negative one. So to answer this question, let’s start by recalling what we mean by the general equation of a plane. We know that the general equation of a plane is of the form 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 is equal to zero, where 𝑎, 𝑏, 𝑐, and 𝑑 are constants. And in particular, the vector 𝑎, 𝑏, 𝑐 will be a normal vector to the plane.
Therefore, to find the general equation of the plane, we’re going to need to find a normal vector to this plane. And we can do this by noting we’re given two parallel vectors to the plane. And since these two vectors are not scalar multiples of each other, we can determine a vector normal to both of these vectors by using the cross product. And of course, if a vector is normal to two vectors parallel to the plane, then that vector will also be a normal of the plane. So we’ll set our vector 𝐧 equal to the cross product of these two vectors. And we recall we find the cross product of two vectors in three dimensions by finding the determinant of a three-by-three matrix.
In the first row of this matrix, we wrote the unit directional vectors 𝐢, 𝐣, and 𝐤. And then in the second and third row, we write the components of the two vectors we’re finding the cross product of. We need to find the determinant of the matrix 𝐢, 𝐣, 𝐤, nine, seven, negative eight, negative two, two, negative one. We can then evaluate this determinant by expanding over the first row. We get 𝐢 times the determinant of the matrix seven, negative eight, two, negative one minus 𝐣 times the determinant of the matrix nine, negative eight, negative two, negative one plus 𝐤 times the determinant of the matrix nine, seven, negative two, two.
Now we just need to evaluate the determinant of each two-by-two matrix. And we recall we do this by finding the difference in the product of the diagonals. Evaluating these determinants gives us 𝐢 times negative seven plus 16 minus 𝐣 multiplied by negative nine minus 16 plus 𝐤 times 18 plus 14. We can then evaluate and simplify this expression. We get nine 𝐢 plus 25𝐣 plus 32𝐤. And any nonzero multiple of this vector is a normal vector to the plane. So we found our values of 𝑎, 𝑏, and 𝑐.
However, we still need to determine the value of 𝑑. We can do this by using the fact that the point with coordinates five, one, negative one lies on the plane. We’ll do this by first substituting our values of 𝑎, 𝑏, and 𝑐 into the general equation of our plane. We get nine 𝑥 plus 25𝑦 plus 32𝑧 plus 𝑑 must be equal to zero. Now, if the point five, one, negative one lies on the plane, it must satisfy the equation of the plane. So we can substitute its coordinates into this equation. This then gives us nine times five plus 25 multiplied by one plus 32 times negative one plus 𝑑 is equal to zero.
We can then evaluate this and solve this equation for 𝑑. We get 45 plus 25 minus 32 plus 𝑑 is equal to zero, which simplifies to give us 38 plus 𝑑 is equal to zero. We subtract 38 from both sides of the equation to see 𝑑 is negative 38. And now we just substitute our values for 𝑎, 𝑏, 𝑐, and 𝑑 into the general equation of the plane. This then gives us our final answer. The general equation of the plane which passes through the point five, one, negative one and is parallel to the two vectors nine, seven, negative eight and negative two, two, negative one is the equation nine 𝑥 plus 25𝑦 plus 32𝑧 minus 38 is equal to zero.