Video Transcript
Given that the plane three 𝑥 minus
three 𝑦 minus three 𝑧 equals one is perpendicular to the plane 𝑎𝑥 minus two 𝑦
minus 𝑧 equals four, find the value of 𝑎.
Okay, in this exercise, we’re given
equations of two planes and told that these planes are perpendicular to one
another. And we can say that if two planes
are perpendicular like these are, then it must also be the case that their normal
vectors are perpendicular. And this points us to the
mathematical condition that is satisfied for two perpendicular planes. The dot product of the normal
vectors of these planes must be zero. This suggests that we might solve
for the normal vectors of our two given planes and then apply this condition to see
if it helps us solve for 𝑎. Since both of these equations are
given in standard form, we can identify the components of each plane’s normal
vector.
For the first plane, its normal
vector has an 𝑥-component of three, a 𝑦-component of negative three, and a
𝑧-component also of negative three. We can write that like this,
calling this normal vector 𝐧 one. For our second plane, its normal
vector has an 𝑥-component of 𝑎, a 𝑦-component of negative two, and a 𝑧-component
of negative one. And we’ll call this normal vector
𝐧 two. Next, knowing that these two planes
are perpendicular, we’ll apply this condition. We’ll say that 𝐧 one, this vector
here, dotted with 𝐧 two, this vector here, equals zero. If we then carry out this dot
product operation, we find that three times 𝑎 plus six plus three equals zero. And this means that three times the
unknown value 𝑎 is equal to negative nine. Dividing both sides by three, we
find that 𝑎 is equal to negative three. This is the value of 𝑎.
