Topic
For triangle ABC square at A, \(AB = 4a,AC = 5a\). Calculate
a) \(\left| {\overrightarrow {AB} – \overrightarrow {AC} } \right|\)
b) \(\left| {\overrightarrow {AB} + \overrightarrow {AC} } \right|\)
Solution method – See details
Step 1: Transform the difference/sum of 2 corresponding vectors into a vector whose prices are the edges ofABC
Step 2: Calculate the lengths of the sides and then deduce the corresponding vector length
Step 3: Build a rectangle ABDCuse the parallelogram rule to calculate the length of the sum of 2 common vectors of origin
Detailed explanation
∆ABC square at A\(AB = 4a,AC = 5a\) \( \Rightarrow BC = \sqrt {A{B^2} + A{C^2}} = a\sqrt {41} \)
a) We have: \(\left| {\overrightarrow {AB} – \overrightarrow {AC} } \right| = \left| {\overrightarrow {CB} } \right| = BC = a\sqrt {41} \ )
b) Construct a rectangle ABCD. Then \(AD = BC = a\sqrt {41} \)
We have: \(\left| {\overrightarrow {AB} + \overrightarrow {AC} } \right| = \left| {\overrightarrow {AD} } \right| = AD = a\sqrt {41} \)