Given the information in the diagram below, find the value of 𝑛 such that vector 𝐀𝐃 plus vector 𝐃𝐄 is equal to 𝑛 multiplied by the vector 𝐀𝐂.
In this question, we will need to use our knowledge of vectors and also similar triangles. Angles 𝐴𝐸𝐷 and 𝐴𝐶𝐵 are corresponding or F angles. This means that they are equal. Likewise, angle 𝐴𝐷𝐸 is equal to angle 𝐴𝐵𝐶. This means that triangle 𝐴𝐷𝐸 is similar to triangle 𝐴𝐵𝐶. The angle at vertex 𝐴 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, 𝐴𝐵 over 𝐴𝐷 is equal to 𝐴𝐶 over 𝐴𝐸. Substituting in the lengths from the diagram, we have 15 over 7.5 is equal to 𝐴𝐶 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 𝐴𝐶. The length of 𝐴𝐶 is, therefore, equal to 21 centimeters. If 𝐴𝐶 is equal to 21 centimeters, then 𝐸𝐶 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 𝐃𝐄 is equal to vector 𝐃𝐀 plus vector 𝐀𝐄. This is because we can get from vertex 𝐷 to vertex 𝐸 via vertex 𝐴. Substituting this into our initial equation gives us vector 𝐀𝐃 plus vector 𝐃𝐀 plus vector 𝐀𝐄 is equal to 𝑛 multiplied by vector 𝐀𝐂. As any vector has both magnitude and direction, the vector 𝐃𝐀 is equal to the negative of vector 𝐀𝐃.
We now have 𝐀𝐃 plus negative 𝐀𝐃 plus 𝐀𝐄 is equal to 𝑛 multiplied by 𝐀𝐂. 𝐀𝐃 plus negative 𝐀𝐃 is equal to zero. So, we have vector 𝐀𝐄 is equal to 𝑛 multiplied by vector 𝐀𝐂. We know that vector 𝐀𝐄 has a magnitude of 10.5 and vector 𝐀𝐂 has a magnitude of 21. We can then divide both sides of this equation by 21, giving us 𝑛 is equal to one-half. This ties in with the fact that the lengths of the sides in triangle 𝐴𝐷𝐸 are half the size of those in 𝐴𝐵𝐶.
The scale factor to go from the larger triangle to the smaller triangle is one-half.