Question

Given the vectors 𝐴⃗=5𝑖̂+3𝑗̂+𝑘̂ and 𝐵⃗⃗=−3𝑖̂+2𝑗̂−4𝑘̂, find the following:

• 𝐴⃗ ∙ 𝐵⃗

• Angle between vector A and B

• 𝐴⃗ 𝑥 𝐵⃗

Answer 1

(a)

$${\vec A}\cdot {\vec B}=(5𝑖̂+3𝑗̂+𝑘̂)\cdot(−3𝑖̂+2𝑗̂−4𝑘̂)\\ =-15+6-4=-13$$

(b)

$$\theta=\cos^{-1}\frac{-13}{\sqrt{5^2+3^2+1^2}\sqrt{(-3)^2+2^2+(-4)^2}}\\ =114^\circ$$

(c)

$${\vec A}\times {\vec B}=(5𝑖̂+3𝑗̂+𝑘̂)\times(−3𝑖̂+2𝑗̂−4𝑘̂)\\ =10𝑘̂+20𝑗̂+9𝑘̂-12𝑖̂-3𝑗̂-2𝑖̂\\ =-14𝑖̂+17𝑗̂+19𝑘̂$$