# Question Video: Finding the Projection of a Vector in the Direction of Another Represented in a Triangle Mathematics • 12th Grade

π΄π΅πΆ is a right triangle at π΅, where π΄π΅ = 17 cm, π΅πΆ = 11 cm, and π· is the midpoint of π΄πΆ. Find the algebraic projection of π΄π· in the direction of πΆπ΅.

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### Video Transcript

π΄π΅πΆ is a right triangle at π΅, where π΄π΅ equals 17 centimeters, π΅πΆ equals 11 centimeters, and π· is the midpoint of π΄πΆ. Find the algebraic projection of π΄π· in the direction of πΆπ΅.

Okay, in this example, we have this right triangle that looks something like this. Weβre told that the length of side π΄π΅ is 17 centimeters, here weβll leave off the units, while π΅πΆ is 11 centimeters. And thereβs a point called point π·, which is midway between π΄ and πΆ on the hypotenuse of the triangle. Our question wants us to solve for the algebraic projection of a vector ππ in the direction of another vector ππ.

First, letβs define ππ. This is a vector that starts at point π΄ and ends at point π·. And likewise, ππ is a vector that starts at point πΆ 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, letβs first solve for the components of these two vectors. Letβs say that point π΅ in our triangle is the origin of an π₯π¦-coordinate frame. We see then that vector ππ 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΅, πΆ, 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 πΆ has coordinates 11, zero. But now what about the coordinates of point π·? Because point π· cuts in half the line segment π΄πΆ, 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 ππ and ππ. Vector ππ 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 ππ.

Next, letβs calculate the components of ππ. To do this, weβll subtract the coordinates of point πΆ from those at point π΅, which means weβ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 ππ in the direction of ππ.

To solve for this, letβs 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 ππ dot ππ divided by the magnitude of ππ. 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 weβre left with negative 5.5. This is the algebraic projection of ππ in the direction of ππ.