Question Video: Finding the Scalar Multiplied by a Vector Using a Given Figure Mathematics

Given the information in the diagram below, if 𝐁𝐂 = π‘˜π„πƒ, find π‘˜.


Video Transcript

Given the information in the diagram below, if the vector 𝐁𝐂 is equal to π‘˜ times the vector 𝐄𝐃, find π‘˜.

In our diagram, we’ve actually been given two similar triangles. That is, one is an enlargement of the other. Triangle 𝐀𝐃𝐄 has been enlarged onto 𝐀𝐁𝐂. Now, we know this since angle 𝐴 here is a shared angle. We see that angle 𝐴𝐸𝐷 and 𝐴𝐢𝐡 are equal as are angles 𝐴𝐷𝐸 and 𝐴𝐡𝐢. And this is because corresponding angles are equal and we know sides 𝐸𝐷 and 𝐢𝐡 are parallel.

Since the triangles have equal angles, they must be similar. And this means we’re able to calculate a scale factor of enlargement. This is found by dividing the length of the enlarged triangle by the corresponding length of the original. We sometimes write this as new length divided by old length.

If we take the enlarged triangle to be 𝐴𝐡𝐢 and the original triangle to be 𝐴𝐷𝐸, we see that we can find the scale factor by dividing the length 𝐴𝐡 by the length 𝐴𝐷. The length of 𝐴𝐡 is actually the sum of the two dimensions given. It’s 7.8 plus 5.2, which is 13 centimeters. And so the scale factor for enlargement here is 13 divided by 5.2, which is five over two.

Now, the scale factor is simply a multiplier. We know that to enlarge triangle 𝐴𝐷𝐸 onto 𝐴𝐡𝐢, we’d multiply any of its lengths by the scale factor of five over two. Now, we’re trying to find a relationship between the vectors 𝐁𝐂 and 𝐄𝐃. Well, we know that the dimension 𝐢𝐡 must be five over two times the dimension 𝐸𝐷. But since these lines are also parallel, we know that they have the same direction. And this means we can say that the vector 𝐂𝐁 must be five over two times the vector 𝐄𝐃.

The problem is, we want to work out the vector 𝐁𝐂 in terms of the vector 𝐄𝐃. Currently, we have the vector 𝐂𝐁 in terms of 𝐄𝐃. And so we’re traveling in the opposite direction. And we remember that to do this with vectors, we change the sign so that the vector 𝐁𝐂 is equal to the negative vector 𝐂𝐁. We just showed that the vector 𝐂𝐁 is five over two times the vector 𝐄𝐃. So this must mean that the vector 𝐁𝐂 is negative five over two times the vector 𝐄𝐃. Comparing this to the original form in the question, and we see then that π‘˜ is equal to negative five over two.

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