Porisma

Worked solution

Expand dot products using magnitudes and an angle

Given ∣a∣=4|\mathbf a|=4, ∣b∣=5|\mathbf b|=5 and the angle between a\mathbf a and b\mathbf b is 45∘45^\circ, find the exact values of the following:

(i) a⋅b\mathbf a\cdot\mathbf b

(ii) (4a+2b)⋅(3a−b)(4\mathbf a+2\mathbf b)\cdot(3\mathbf a-\mathbf b)

(iii) ∣a−b∣2|\mathbf a-\mathbf b|^2

CJC · Discussion · Question 5

Answer

(i) 10210\sqrt 2; (ii) 142+202142+20\sqrt 2; (iii) 41−20241-20\sqrt 2.

(i) Use the scalar-product angle formula

a⋅b=∣a∣∣b∣cos⁡45∘=4(5)(2/2)=102\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos 45^\circ=4(5)(\sqrt 2/2)=10\sqrt 2.

How did this part go?

(ii) Distribute the dot product

(4a+2b)⋅(3a−b)=12a⋅a−4a⋅b+6b⋅a−2b⋅b=12∣a∣2+2a⋅b−2∣b∣2=12(16)+202−2(25)=142+202.\begin{aligned}(4\mathbf a+2\mathbf b)\cdot(3\mathbf a-\mathbf b)&=12\mathbf a\cdot\mathbf a-4\mathbf a\cdot\mathbf b+6\mathbf b\cdot\mathbf a-2\mathbf b\cdot\mathbf b\\&=12|\mathbf a|^2+2\mathbf a\cdot\mathbf b-2|\mathbf b|^2\\&=12(16)+20\sqrt 2-2(25)\\&=142+20\sqrt 2.\end{aligned}

How did this part go?

(iii) Square the magnitude through a dot product

∣a−b∣2=(a−b)⋅(a−b)=∣a∣2−2a⋅b+∣b∣2=16−202+25=41−202.\begin{aligned}|\mathbf a-\mathbf b|^2&=(\mathbf a-\mathbf b)\cdot(\mathbf a-\mathbf b)\\&=|\mathbf a|^2-2\mathbf a\cdot\mathbf b+|\mathbf b|^2\\&=16-20\sqrt 2+25\\&=41-20\sqrt 2.\end{aligned}

How did this part go?

Common mistakes

Part (iii) asks for the squared magnitude, not the vector difference or its unsquared length.

Solution notes

All three asks and the printed expansions retained. This discussion sheet prints no mark allocations for Questions 1–11.

Read the hint