Video Transcript
π΄π΅πΆ is a right triangle at π΅,
where π΄π΅ equals 17 centimeters, π΅πΆ equals 11 centimeters, and π· is the midpoint
of π΄πΆ. Find the algebraic projection of
π΄π· in the direction of πΆπ΅.
Okay, in this example, we have this
right triangle that looks something like this. Weβre told that the length of side
π΄π΅ is 17 centimeters, here weβll leave off the units, while π΅πΆ is 11
centimeters. And thereβs a point called point
π·, which is midway between π΄ and πΆ on the hypotenuse of the triangle. Our question wants us to solve for
the algebraic projection of a vector ππ in the direction of another vector
ππ.
First, letβs define ππ. This is a vector that starts at
point π΄ and ends at point π·. And likewise, ππ is a vector that
starts at point πΆ and ends at point π΅. So the idea is, we want to solve
for the algebraic projection of this vector onto this one.
To begin figuring this out, letβs
first solve for the components of these two vectors. Letβs say that point π΅ in our
triangle is the origin of an π₯π¦-coordinate frame. We see then that vector ππ lies
along the π₯-axis and line segment π΄π΅ lies along the π¦. From this perspective, we can
define the coordinates of the four points π΄, π΅, πΆ, and π·.
Point π΄ has an π₯-coordinate of
zero and a π¦-coordinate of 17. Point π΅, because it exists at the
origin, has coordinates zero, zero, while point πΆ has coordinates 11, zero. But now what about the coordinates
of point π·? Because point π· cuts in half the
line segment π΄πΆ, that means its π₯- and π¦-coordinates are one-half of the side
lengths of those two sides of our triangle. That is, the π₯-coordinate of π· is
11 divided by two or 5.5, while the π¦-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 ππ and ππ. Vector ππ is equal to the
coordinates of point π· minus those of point π΄ all in vector form. When we substitute in for the
coordinates of points π· and π΄, we find that this subtraction gives us a result of
5.5 and negative 8.5. These then are the π₯- and
π¦-components of vector ππ.
Next, letβs calculate the
components of ππ. To do this, weβll subtract the
coordinates of point πΆ from those at point π΅, which means weβll 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 ππ in the direction
of ππ.
To solve for this, letβs remember
that the scalar, also known as the algebraic, projection of one vector, π one, onto
another, π 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 ππ dot ππ divided by the magnitude of ππ. 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 weβre left with negative
5.5. This is the algebraic projection of
ππ in the direction of ππ.