Video Transcript
Given that the plane two 𝑥 plus
six 𝑦 plus two 𝑧 equals 18 intersects the coordinate axes 𝑥, 𝑦, and 𝑧 at the
points 𝐴, 𝐵, and 𝐶, respectively, find the area of triangle 𝐴𝐵𝐶.
Okay, so in this example, we have a
plane that we’re told intersects the 𝑥-, 𝑦-, and 𝑧-axes. And we’re told something about the
points of intersection. We know, for instance, that the
coordinates of the point of intersection of our plane and the 𝑥-axis are capital
𝐴, zero, and zero. Similarly, the intersection of the
𝑦-axis in our plane happens at the point zero, capital 𝐵, zero. And along the 𝑧-axis, it occurs at
zero, zero, capital 𝐶.
This tells us that if we were to
write the equation of our plane in what’s called intercept form, it would look like
this: 𝑥 over capital 𝐴 plus 𝑦 over capital 𝐵 plus 𝑧 over capital 𝐶 equals
one. And note that we are given an
equation for our plane, but it’s just not yet in intercept form.
We’re interested though in
rearranging it so it is in this form, and here’s why. Going back to our sketch, we see
that our three points of intersection define the vertices of this triangle
𝐴𝐵𝐶. If we can solve for the actual
values capital 𝐴, capital 𝐵, and capital 𝐶, then we can make progress in solving
for the area of our triangle. With that goal in mind, let’s take
this given form of our plane’s equation and work to convert it to intercept
form.
To do that, let’s notice that in
intercept form our equation will have one on one of its sides. In the current given form, we have
18, but we can rearrange that so it equals one by dividing both sides of our
equation by 18. If we distribute the 18 in the
denominator on the left, we end up with 𝑥 over nine plus 𝑦 over three plus 𝑧 over
nine equaling one. And this is wonderful because now
our equation is in intercept form. And so we see that capital 𝐴
equals nine, capital 𝐵 is equal to three, and capital 𝐶 also equals nine.
Now that we know the coordinates of
our triangle’s vertices, we can use this information to solve for its area. To help us do that, we’ll define
two vectors that form sides of this triangle. The first vector, what we’ve called
𝐯 one, is equal to our 𝑥-axis intercept minus our 𝑦-axis intercept. This vector then has components
nine, negative three, zero. We can then define a second vector
along the side of our triangle. We’ll call it 𝐯 two. And this is equal to our 𝑥-axis
intercept minus our 𝑧-axis intercept. Therefore, 𝐯 two has components
nine, zero, negative nine.
Now that we have two vectors which
define sides in our triangle, let’s recall that given two vectors 𝐯 one and 𝐯 two
that define adjacent sides of a parallelogram, then the total area of that
parallelogram is equal to the magnitude of the cross product of 𝐯 one and 𝐯
two. This fact is actually helpful to us
here because if we join the points of vectors 𝐯 one and 𝐯 two, then we see we have
a triangle whose area is equal to one-half the area of this parallelogram. That is, the area of the triangle
with sides 𝐯 one and 𝐯 two is equal to one-half the magnitude of the cross product
of these vectors.
It’s this relationship that we’ll
use to solve for the area of triangle 𝐴𝐵𝐶. We’ll move this equation off to the
side. And to calculate the area of
triangle 𝐴𝐵𝐶, let’s first take the cross product of 𝐯 one and 𝐯 two. That’s equal to the determinant of
this three-by-three matrix, where we see the first row of the matrix is populated by
the 𝐢, 𝐣, and 𝐤 unit vectors and the second and third rows have the correspondent
components of 𝐯 one and 𝐯 two. This cross product equals 𝐢 hat
times 27 minus 𝐣 hat times negative 81 plus 𝐤 hat times 27. Written in slightly different form,
we can express this as 27, 81, 27.
Let’s remember now we want to take
the magnitude of this cross product to help us solve for the area of our
triangle. That is, we want to solve for the
magnitude of the vector with components 27, 81, 27. This is equal to the square root of
the sum of the squares of the different components of this vector. As we simplify what’s inside the
square root sign, we note that 81 can be written as three times 27. This means we have 27 squared in
each one of the three terms under the square root, which tells us we can factor it
completely outside. One plus three squared plus one is
11. So our result is 27 times the
square root of 11.
This though is not quite our answer
for the area of this triangle because, remember, there’s a factor of one-half we
need to multiply this result by. Our final answer then is that the
area of triangle 𝐴𝐵𝐶 is equal to 27 over two times the square root of 11.
