Question Video: Finding the Projection of a Vector in the Direction of Another Represented in a Square

Given that 𝐴𝐡𝐢𝐷 is a square having a side length of 53 cm, calculate the projection of π‚πš¨ in the direction of πš©π‚.

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Video Transcript

Given that 𝐴𝐡𝐢𝐷 is a square having a side length of 53 centimeters, calculate the projection of π‚πš¨ in the direction of πš©π‚.

Okay, let’s say that this is our square 𝐴𝐡𝐢𝐷. And we know that the length of each side of the square is 53 centimeters. Here we’ll leave off the units. We want to calculate the projection of a vector from point 𝐢 to point 𝐴 in the direction of a vector from point 𝐡 to point 𝐢. So first, vector π‚πš¨ would look like this pointing from point 𝐢 to point 𝐴, and then vector πš©π‚ would look like this. We want to calculate then the projection of this vector in the direction of this one. To figure this out, we’ll need to know the components of these two vectors. Vector π‚πš¨ is equal to the coordinates of point 𝐴 minus those of point 𝐂 all in vector form.

Let’s say that point 𝐴 in our square is at the origin of an π‘₯𝑦-coordinate frame. The π‘₯-axis goes out along line 𝐴𝐡, and the 𝑦-axis along line 𝐴𝐷. Using this as a reference, we can solve for the coordinates of points 𝐴 and 𝐢. Since 𝐴 is at the origin, its coordinates are zero, zero and 𝐢 has coordinates of 53 and 53. Therefore, this vector overall has components negative 53, negative 53. Next, we’ll solve for the components of vector πš©π‚. This is equal to the vector form of the coordinates of point 𝐢 minus those of point 𝐡. We’ve already seen that the coordinates of point 𝐢 are 53, 53. And looking at our sketch, we can see that those of point 𝚩 are 53 in the π‘₯-direction and zero in the 𝑦-direction. Computing this difference, we end up with the vector with components zero, positive 53.

Now that we have the components of our two vectors of interest, let’s recall that, in general, this scalar projection of one vector onto another is equal to the dot product of those vectors divided by the magnitude of the vector being projected onto. This tells us that the projection of π‚πš¨ in the direction of πš©π‚ is equal to π‚πš¨ dot πš©π‚ over the magnitude of πš©π‚. Plugging in for the components of π‚πš¨ and πš©π‚, we get this expression where we’ve recalled that the magnitude of a given vector equals the square root of the sum of the squares of the components of that vector.

In computing this fraction, our first step in the numerator is to begin calculating this dot product by multiplying together the corresponding components of these vectors. And then in the denominator, we can square these components of vector πš©π‚. In our numerator, we find a result of negative quantity 53 squared. And in our denominator, we have a square root of 53 squared, which means that our denominator simplifies to 53. And a factor of 53 cancels from top and bottom so that our final result is negative 53.

This is the projection of vector π‚πš¨ in the direction of πš©π‚. And we see this confirmed graphically since it is this length here which equals this projection. That’s the side length of our square. And then because along this line the projection of π‚πš¨ and πš©π‚ act in opposite directions, our final result is negative.

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