### Video Transcript

Which of the following is the
smaller angle between the straight line 𝐫 equals negative seven 𝐢 minus 𝐣 minus
nine 𝐤 plus 𝑡 times two 𝐢 plus 𝐣 minus 𝐤 and the plane 𝐫 dot nine 𝐢 minus
nine 𝐣 plus two 𝐤 equals 13? (A) The inverse sin of seven times
the square root of 249 divided by 498. (B) The inverse cos of seven times
the square root of 249 divided by 498. (C) The inverse sin of seven times
the square root of 43 over 86. (D) The inverse cos of seven times
the square root of 43 over 86.

Okay, in this exercise, we’re given
the equation of a line and the equation of a plane. Let’s imagine that this is our line
and this is our plane. We want to solve for the smaller
angle between these two objects. We imagine this line and plane
intersecting in three-dimensional space. And we see that that intersection
forms two angles, this one here as well as this one. Our problem statement tells us to
solve for the smaller of these two angles.

To do this, we’ll need to know a
vector that is normal or perpendicular to our plane as well as a vector that is
parallel to our line. Beginning with our plane, let’s say
that a vector normal to its surface is called 𝐧. If we look at the form in which the
equation of our plane is given, we see it’s presented to us in what’s called vector
form. Written this way, a vector to a
general point in the plane is dotted with a vector normal to the plane. From this then, we can read off the
components of our vector 𝐧, nine, negative nine, two.

Now let’s look to solve for a
vector that is parallel to our line. We’ll call this vector 𝐩. If we look at the form in which the
equation of our line is given, we can find a vector that’s parallel to this line by
looking for the scale factor; in this case it’s called 𝑡. The vector that multiplies the
scale factor is parallel to our line. We can say then that the components
of 𝐩 are two, one, and negative one. Now that we have our vector normal
to our plane and a vector parallel to our line, we can recall the general equation
that the sin of the angle 𝛼 between a plane and a line is equal to the magnitude of
the dot product of a vector parallel to the line and one normal to the plane all
divided by the product of the magnitude of each of these vectors.

To solve for 𝛼 then, we can go
about calculating this numerator and denominator and then bringing them
together. The magnitude of 𝐩 dot 𝐧 using
our 𝐩 and 𝐧 vectors equals the magnitude of two, one, negative one dotted with
nine, negative nine, two. This equals the magnitude of 18
minus nine minus two or seven. Knowing this result, now let’s
calculate the denominator of our fraction. This equals the square root of the
sum of the squares of two, one, and negative one times the square root of the sum of
the squares of nine, negative nine, and two. That equals this expression or the
square root of six times the square root of 166, which equals the square root of
996.

We can now say that the sin of the
angle 𝛼 equals seven over the square root of 996. And if we rationalize our
denominator, this equals seven times the square root of 996 over 996. And then we note that 996 is equal
to four times 249. Written this way, we can bring the
four outside of the square root, where it becomes a two. It turns out, though, that 996 is
divisible by two. So if we divide numerator and
denominator of this fraction by two, we find it equals seven times the square root
of 249 divided by 498. So if the sin of 𝛼 is equal to
this expression, then 𝛼 is the inverse sin of seven times the square root of 249
over 498.

Now, if we evaluate this on our
calculator, we find a result of about 13 degrees. This confirms to us that, indeed,
we’ve solved for the smaller of the two angles of intersection between our line and
the plane. Our result matches up with answer
option (A). The smaller angle between this line
and plane is the inverse sin of seven times the square root of 249 divided by
498.