Question Video: Finding the Length of the Projection of a Vector on Another Vector Mathematics • Third Year of Secondary School

Video Transcript

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

Alright, looking at this cube, we see that corner 𝑂 is at the origin of an 𝑥𝑦𝑧-coordinate system. We also see three other corners of the cube 𝐴, 𝐵, and 𝐶 labeled. We want to solve for the scalar projection of the vector 𝐎𝐀 shown in our sketch onto the vector 𝐂𝐁 which goes from corner 𝐶 of our cube to corner 𝐵. That vector would look like this. To start working toward our solution, let’s recall that the scalar projection of a vector 𝐀 onto another vector 𝐁 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 𝐎𝐀 dotted with 𝐂𝐁 divided by the magnitude of vector 𝐂𝐁.

To calculate this fraction, we’re going to need to know the vector components of these two vectors, 𝐎𝐀 and 𝐂𝐁. We can find them out by working with the coordinates of our four given points. We can begin with the coordinates of our point 𝑂, which we see is at the origin. Because of that, we know the 𝑥𝑦𝑧-coordinates of this point must be zero, zero, zero.

Next, let’s consider the coordinates of point 𝐴. That point has an 𝑥-value that’s equal to this length, a length of the side of the cube, which we’re told equals forty-four seventeenths, a 𝑦-value that’s equal to this length, also one side length of the cube, and then a 𝑧-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 𝐎𝐀. That vector equals the coordinates of point 𝐴 minus those of point 𝑂. As we’ve seen, the 𝑥-, 𝑦-, and 𝑧-coordinates of 𝐴 are all forty-four seventeenths, and those of point 𝑂 are all zero. This tells us that the components of vector 𝐎𝐀 are also all the same. They’re all forty-four seventeenths. Knowing this, now let’s figure out the coordinates of the other two points 𝐶 and 𝐵. Starting with point 𝐵, we can see that this has an 𝑥-coordinate of zero, a 𝑦-coordinate of zero, but then a 𝑧-coordinate of forty-four seventeenths. Point 𝐶 then is in some sense the opposite. It has 𝑥- and 𝑦-values of 44 over 17 and a 𝑧-value of zero.

Vector 𝐂𝐁 equals the vector form of the coordinates of 𝐵 minus those of 𝐶. And when we substitute in these values and perform this subtraction, we find this result. An 𝑥-component of negative forty-four seventeenths, a 𝑦-component of the same, and a 𝑧-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, let’s clear a bit of space on screen. And now let’s substitute in the known values for 𝐎𝐀 and 𝐂𝐁 into this fraction. That gives us this expression, where we’ve 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 𝑥-, 𝑦-, and 𝑧-directions. In our denominator, we can square these various components of vector 𝐂𝐁. 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 𝐎𝐀 onto vector 𝐂𝐁 in our cube.