Video Transcript
Determine whether the planes 𝑥
plus three 𝑦 plus four 𝑧 equals six and 𝑥 over five plus three 𝑦 over five plus
four 𝑧 over five equals one are parallel or perpendicular.
Okay, so here we have these two
plane equations, and they’re given to us nearly in the form that’s called general
form. When a plane is given to us in
what’s called general form, then we can take the values by which we multiply 𝑥, 𝑦,
and 𝑧. And these values form the
components of a vector that is normal to the plane. Now, if we were to call this first
equation given to us that of plane one, we could subtract six from both sides of the
equation and get this result. Notice that now this equation is in
general form. And so if we call a vector normal
to this plane 𝐧 one, we can say that the components of this vector are positive
one, positive three, and positive four.
We’re also interested in solving
for a vector normal to our second plane. We’ll call it plane two. The reason for this is if the
vectors that are normal to our two planes themselves are parallel or perpendicular,
then so are the planes. That being said, considering our
second plane’s equation, if we subtract one from both sides, we get this plane’s
equation in general form. And from this, we can solve for a
vector 𝐧 two normal to this second plane. It has components one-fifth,
positive three-fifths, and positive four-fifths.
Now that we have vectors normal to
each plane, we can start to test for being parallel or perpendicular. Let’s first figure out whether
these two planes are parallel to one another. If they are, then, as we mentioned,
their normal vectors will be parallel too. And that means that there exists
some constant, we’ll call it 𝐶, by which we can multiply 𝐧 two to give 𝐧 one. To figure out whether our planes
are parallel then, we’ll want to solve and find if some such 𝐶 exists.
We can begin doing this by
comparing the 𝑥-components of our two normal vectors. The question we want to answer is,
by what value do we need to multiply this component so that it equals this one? In other words, if one is equal to
a constant 𝐶 times one-fifth, then what is 𝐶? Solving this equation for 𝐶, we
find that it’s equal to five. This is what we could call the
constant of proportionality for the 𝑥-components of these two normal vectors. And then we test the 𝑦 and 𝑧 to
see if the same relationship holds.
Considering the 𝑦-components, if
we multiply three-fifths by our value for 𝐶, do we get three? Well, if 𝐶 is five, then five
times three-fifths is indeed three. And then, finally, if we multiply
this value four-fifths by 𝐶, do we get four? We can tell that we will; with 𝐶
being equal to five, five times four-fifths is four. So then we found a constant
multiple by which these two normal vectors relate to one another. Since they can be expressed this
way, that means the two normal vectors are parallel. And therefore, planes one and two
are as well. Our final answer then is that these
two planes are parallel to one another.
