3.2 Q-16 | Vectors | Mathematics 11

Note: The negative sign indicates that the projection points in the direction opposite to the reference vector, since the angle between $a$ and $b$ is obtuse ( $cos θ < 0$ ).

Recall: Dot Product in Component Form

For $u = u_{1} \hat{i} + u_{2} \hat{j} + u_{3} \hat{k}$ and $v = v_{1} \hat{i} + v_{2} \hat{j} + v_{3} \hat{k}$ :

$u \cdot v = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}$

Geometrically: $a \cdot b = ∣ a ∣∣ b ∣ cos θ$

Exercise 3.2 — Question 16

Problem

Find the projection of $a$ along $b$ and the projection of $b$ along $a$ , given:

$a = 2 \hat{i} - 3 \hat{j} + \hat{k}, b = \hat{i} + 2 \hat{j} - 4 \hat{k}$

Key Formula

The scalar projection of $a$ along $b$ is:

$proj_{b} a = \frac{a \cdot b}{∣ b ∣}$

The scalar projection of $b$ along $a$ is:

$proj_{a} b = \frac{a \cdot b}{∣ a ∣}$

Step 1 — Compute the Dot Product

Using the component form $u \cdot v = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}$ :

$a \cdot b = (2) (1) + (- 3) (2) + (1) (- 4)$ $= 2 - 6 - 4 = - 8$

Step 2 — Compute the Magnitudes

$∣ a ∣ = (2)^{2} + (- 3)^{2} + (1)^{2} = 4 + 9 + 1 = 14$

$∣ b ∣ = (1)^{2} + (2)^{2} + (- 4)^{2} = 1 + 4 + 16 = 21$

Step 3 — Compute the Projections

Projection of $a$ along $b$ :

$proj_{b} a = \frac{a \cdot b}{∣ b ∣} = \frac{- 8}{21}$

Projection of $b$ along $a$ :

$proj_{a} b = \frac{a \cdot b}{∣ a ∣} = \frac{- 8}{14}$

Note: The negative sign indicates that the projection points in the direction opposite to the reference vector, since the angle between $a$ and $b$ is obtuse ( $cos θ < 0$ ).

Recall: Dot Product in Component Form

For $u = u_{1} \hat{i} + u_{2} \hat{j} + u_{3} \hat{k}$ and $v = v_{1} \hat{i} + v_{2} \hat{j} + v_{3} \hat{k}$ :

$u \cdot v = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}$

Geometrically: $a \cdot b = ∣ a ∣∣ b ∣ cos θ$