### Video Transcript

The cube shown has sides of length 44 divided by 17. Find the scalar projection of ππ onto ππ giving your answer correct to two decimal places.

Alright, looking at this cube, we see that corner π is at the origin of an π₯π¦π§-coordinate system. We also see three other corners of the cube π΄, π΅, and πΆ labeled. We want to solve for the scalar projection of the vector ππ shown in our sketch onto the vector ππ which goes from corner πΆ of our cube to corner π΅. That vector would look like this. To start working toward our solution, letβs recall that the scalar projection of a vector π onto another vector π is equal to the dot product of those vectors divided by the magnitude of the vector being projected onto. In our case then, what we want to calculate is ππ dotted with ππ divided by the magnitude of vector ππ.

To calculate this fraction, weβre going to need to know the vector components of these two vectors, ππ and ππ. We can find them out by working with the coordinates of our four given points. We can begin with the coordinates of our point π, which we see is at the origin. Because of that, we know the π₯π¦π§-coordinates of this point must be zero, zero, zero.

Next, letβs consider the coordinates of point π΄. That point has an π₯-value thatβs equal to this length, a length of the side of the cube, which weβre told equals forty-four seventeenths, a π¦-value thatβs equal to this length, also one side length of the cube, and then a π§-value that is equal to this length, once again, forty-four seventeenths.

Now that we knew the coordinates of these two points, we can solve for the vector ππ. That vector equals the coordinates of point π΄ minus those of point π. As weβve seen, the π₯-, π¦-, and π§-coordinates of π΄ are all forty-four seventeenths, and those of point π are all zero. This tells us that the components of vector ππ are also all the same. Theyβre all forty-four seventeenths. Knowing this, now letβs figure out the coordinates of the other two points πΆ and π΅. Starting with point π΅, we can see that this has an π₯-coordinate of zero, a π¦-coordinate of zero, but then a π§-coordinate of forty-four seventeenths. Point πΆ then is in some sense the opposite. It has π₯- and π¦-values of 44 over 17 and a π§-value of zero.

Vector ππ equals the vector form of the coordinates of π΅ minus those of πΆ. And when we substitute in these values and perform this subtraction, we find this result. An π₯-component of negative forty-four seventeenths, a π¦-component of the same, and a π§-component of positive forty-four seventeenths. So here, we have the components of our vectors of interest, which means we can start to calculate this fraction. As we start that, letβs clear a bit of space on screen. And now letβs substitute in the known values for ππ and ππ into this fraction. That gives us this expression, where weβve recalled that the magnitude of a vector equals the square root of the sum of the squares of its components.

In our numerator, our next step is to multiply together the corresponding components of these vectors in the π₯-, π¦-, and π§-directions. In our denominator, we can square these various components of vector ππ. When we do this, in our numerator, we have negative the quantity forty-four seventeenths squared minus that same quantity squared plus that quantity squared and in our denominator the square root of three times the quantity 44 over 17 squared. In our numerator then, this term cancels with this term and in our denominator, recognizing that the square root of forty-four seventeenths squared is simply forty-four seventeenths, we now have this quantity here. And we see that a factor of 44 over 17 can cancel from top and bottom.

This gives us negative one over root three times forty-four seventeenths. This answer is exact, but we want to report our result to the nearest two decimal places. To that level of precision, this is equal to negative 1.49. This is the projection of vector ππ onto vector ππ in our cube.