**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 of*ABC*

Step 2: Calculate the lengths of the sides and then deduce the corresponding vector length

Step 3: Build a rectangle *ABDC*use 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: \(\left| {\overrightarrow {AB} + \overrightarrow {AC} } \right| = \left| {\overrightarrow {AD} } \right| = AD = a\sqrt {41} \)