Video Transcript
𝐴𝐵𝐶 is a right triangle at 𝐵,
where 𝐴𝐵 equals 17 centimeters, 𝐵𝐶 equals 11 centimeters, and 𝐷 is the midpoint
of 𝐴𝐶. Find the algebraic projection of
𝐴𝐷 in the direction of 𝐶𝐵.
Okay, in this example, we have this
right triangle that looks something like this. We’re told that the length of side
𝐴𝐵 is 17 centimeters, here we’ll leave off the units, while 𝐵𝐶 is 11
centimeters. And there’s a point called point
𝐷, which is midway between 𝐴 and 𝐶 on the hypotenuse of the triangle. Our question wants us to solve for
the algebraic projection of a vector 𝐀𝐃 in the direction of another vector
𝐂𝐁.
First, let’s define 𝐀𝐃. This is a vector that starts at
point 𝐴 and ends at point 𝐷. And likewise, 𝐂𝐁 is a vector that
starts at point 𝐶 and ends at point 𝐵. So the idea is, we want to solve
for the algebraic projection of this vector onto this one.
To begin figuring this out, let’s
first solve for the components of these two vectors. Let’s say that point 𝐵 in our
triangle is the origin of an 𝑥𝑦-coordinate frame. We see then that vector 𝐂𝐁 lies
along the 𝑥-axis and line segment 𝐴𝐵 lies along the 𝑦. From this perspective, we can
define the coordinates of the four points 𝐴, 𝐵, 𝐶, and 𝐷.
Point 𝐴 has an 𝑥-coordinate of
zero and a 𝑦-coordinate of 17. Point 𝐵, because it exists at the
origin, has coordinates zero, zero, while point 𝐶 has coordinates 11, zero. But now what about the coordinates
of point 𝐷? Because point 𝐷 cuts in half the
line segment 𝐴𝐶, that means its 𝑥- and 𝑦-coordinates are one-half of the side
lengths of those two sides of our triangle. That is, the 𝑥-coordinate of 𝐷 is
11 divided by two or 5.5, while the 𝑦-coordinate is 17 divided by two or 8.5.
Knowing all this, we can now focus
on solving for the components of our two vectors 𝐀𝐃 and 𝐂𝐁. Vector 𝐀𝐃 is equal to the
coordinates of point 𝐷 minus those of point 𝐴 all in vector form. When we substitute in for the
coordinates of points 𝐷 and 𝐴, we find that this subtraction gives us a result of
5.5 and negative 8.5. These then are the 𝑥- and
𝑦-components of vector 𝐀𝐃.
Next, let’s calculate the
components of 𝐂𝐁. To do this, we’ll subtract the
coordinates of point 𝐶 from those at point 𝐵, which means we’ll subtract the point
11, zero from the point zero, zero. This results in the vector negative
11, zero. Okay, great, so now we have our two
vectors, and we want to solve for the algebraic projection of 𝐀𝐃 in the direction
of 𝐂𝐁.
To solve for this, let’s remember
that the scalar, also known as the algebraic, projection of one vector, 𝐕 one, onto
another, 𝐕 two, is given by the dot product of these two vectors divided by the
magnitude of the vector being projected onto. In our case then, what we want to
calculate is 𝐀𝐃 dot 𝐂𝐁 divided by the magnitude of 𝐂𝐁. Substituting in for the components
of these vectors and recalling that the magnitude of a vector equals the square root
of the sum of the squares of its components, we next start to compute our dot
product by multiplying together the respective components of our vectors.
Along with this, we notice that
negative 11 squared is 121 and zero squared is zero, which means that this
projection equals 5.5 times negative 11 divided by the square root of 121. But then 121 is 11 squared so that
factors of 11 in numerator and denominator cancel out. And we’re left with negative
5.5. This is the algebraic projection of
𝐀𝐃 in the direction of 𝐂𝐁.
