Question Video: Finding the Scalar That Satisfies a Given Operation on Vectors Represented in a Figure | Nagwa Question Video: Finding the Scalar That Satisfies a Given Operation on Vectors Represented in a Figure | Nagwa

# Question Video: Finding the Scalar That Satisfies a Given Operation on Vectors Represented in a Figure Mathematics • First Year of Secondary School

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Given the information in the diagram below, find the value of π such that ππ + ππ = πππ.

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### Video Transcript

Given the information in the diagram below, find the value of π such that vector ππ plus vector ππ is equal to π multiplied by the vector ππ.

In this question, we will need to use our knowledge of vectors and also similar triangles. Angles π΄πΈπ· and π΄πΆπ΅ are corresponding or F angles. This means that they are equal. Likewise, angle π΄π·πΈ is equal to angle π΄π΅πΆ. This means that triangle π΄π·πΈ is similar to triangle π΄π΅πΆ. 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΄πΆ over π΄πΈ. Substituting in the lengths from the diagram, we have 15 over 7.5 is equal to π΄πΆ 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΄πΆ. The length of π΄πΆ is, therefore, equal to 21 centimeters. If π΄πΆ is equal to 21 centimeters, then πΈπΆ 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 ππ is equal to vector ππ plus vector ππ. This is because we can get from vertex π· to vertex πΈ via vertex π΄. Substituting this into our initial equation gives us vector ππ plus vector ππ plus vector ππ is equal to π multiplied by vector ππ. As any vector has both magnitude and direction, the vector ππ is equal to the negative of vector ππ.

We now have ππ plus negative ππ plus ππ is equal to π multiplied by ππ. ππ plus negative ππ is equal to zero. So, we have vector ππ is equal to π multiplied by vector ππ. We know that vector ππ has a magnitude of 10.5 and vector ππ 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·πΈ are half the size of those in π΄π΅πΆ.

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

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