### Video Transcript

Given the information in the diagram below, find the value of π such that vector ππ plus vector ππ is equal to π multiplied by the vector ππ.

In this question, we will need to use our knowledge of vectors and also similar triangles. Angles π΄πΈπ· and π΄πΆπ΅ are corresponding or F angles. This means that they are equal. Likewise, angle π΄π·πΈ is equal to angle π΄π΅πΆ. This means that triangle π΄π·πΈ is similar to triangle π΄π΅πΆ. The angle at vertex π΄ is in both of these triangles. This means that all three angles in both triangles are the same. The property of three angles being equal means that the triangles are similar.

When dealing with similar triangles, we know that the lengths of corresponding sides have the same ratio. In this case, π΄π΅ over π΄π· is equal to π΄πΆ over π΄πΈ. Substituting in the lengths from the diagram, we have 15 over 7.5 is equal to π΄πΆ over 10.5. 15 divided by 7.5 is equal to two. We can then multiply both sides of this equation by 10.5, giving us 10.5 multiplied by two is equal to π΄πΆ. The length of π΄πΆ is, therefore, equal to 21 centimeters. If π΄πΆ is equal to 21 centimeters, then πΈπΆ must be equal to 10.5 centimeters. This is because 21 minus 10.5 is 10.5.

We will now clear some space and consider the vector equation we were given in the question. We know that vector ππ is equal to vector ππ plus vector ππ. This is because we can get from vertex π· to vertex πΈ via vertex π΄. Substituting this into our initial equation gives us vector ππ plus vector ππ plus vector ππ is equal to π multiplied by vector ππ. As any vector has both magnitude and direction, the vector ππ is equal to the negative of vector ππ.

We now have ππ plus negative ππ plus ππ is equal to π multiplied by ππ. ππ plus negative ππ is equal to zero. So, we have vector ππ is equal to π multiplied by vector ππ. We know that vector ππ has a magnitude of 10.5 and vector ππ has a magnitude of 21. We can then divide both sides of this equation by 21, giving us π is equal to one-half. This ties in with the fact that the lengths of the sides in triangle π΄π·πΈ are half the size of those in π΄π΅πΆ.

The scale factor to go from the larger triangle to the smaller triangle is one-half.