3.2 Q-9 | Vectors | Mathematics 11

Exercise 3.2 — Question 9

Find the angle between the vectors: $a = 2 \hat{i} - \hat{j} + \hat{k} and b = - \hat{i} + \hat{j}$

The angle $θ$ between two vectors $a$ and $b$ is given by:

$cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$

$a = 2 \hat{i} - \hat{j} + \hat{k} ⟹ (a_{1}, a_{2}, a_{3}) = (2, - 1, 1)$ $b = - \hat{i} + \hat{j} + 0 \hat{k} ⟹ (b_{1}, b_{2}, b_{3}) = (- 1, 1, 0)$

$a \cdot b = (2) (- 1) + (- 1) (1) + (1) (0)$ $= - 2 - 1 + 0 = - 3$

$∣ a ∣ = (2)^{2} + (- 1)^{2} + (1)^{2} = 4 + 1 + 1 = 6$

$∣ b ∣ = (- 1)^{2} + (1)^{2} + (0)^{2} = 1 + 1 + 0 = 2$

$cos θ = \frac{- 3}{6 \cdot 2} = \frac{- 3}{12} = \frac{- 3}{2 3} = \frac{- 3}{2}$

$θ = cos^{- 1} (\frac{- 3}{2}) = 150°$

$θ = 150°$