## Solve Lesson 38 Page 92 SBT Math 10 – Kite>

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: $$\left| {\overrightarrow {AB} + \overrightarrow {AC} } \right| = \left| {\overrightarrow {AD} } \right| = AD = a\sqrt {41}$$