3.2 Q-14 | Vectors | Mathematics 11

Exercise 3.2 — Question 14

Problem Statement

Find the value of $k$ such that the vectors $a = 2 \hat{i} + k \hat{j} - \hat{k}$ and $b = 3 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are perpendicular to each other.

Key Concepts

Dot Product in Component Form

For two vectors $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ , the dot product is:

$a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$

Condition for Perpendicularity (Orthogonality)

Two non-zero vectors $a$ and $b$ are perpendicular (orthogonal) if and only if:

$a \cdot b = 0$

This follows from the geometric definition $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ , since $cos 9 0^{\circ} = 0$ .

Solution

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

Step 1: Apply the perpendicularity condition $a \cdot b = 0$ .

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

Step 2: Expand and simplify.

$= 6 - 2 k - 4 = 2 - 2 k$

Step 3: Set equal to zero and solve for $k$ .

$2 - 2 k = 0 ⟹ 2 k = 2 ⟹ k = 1$

Answer: $k = 1$

Verification: With $k = 1$ : $a \cdot b = (2) (3) + (1) (- 2) + (- 1) (4) = 6 - 2 - 4 = 0 ✓$

The angle $θ$ between two vectors can be found using:

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

where $∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$ and $∣ b ∣ = b_{1}^{2} + b_{2}^{2} + b_{3}^{2}$ .

When $a \cdot b = 0$ , we get $cos θ = 0$ , confirming $θ = 9 0^{\circ}$ (perpendicular vectors).

Exercise 3.2 — Question 14

Problem Statement

Find the value of $k$ such that the vectors $a = 2 \hat{i} + k \hat{j} - \hat{k}$ and $b = 3 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are perpendicular to each other.

Key Concepts

Dot Product in Component Form

For two vectors $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ , the dot product is:

$a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$

Condition for Perpendicularity (Orthogonality)

Two non-zero vectors $a$ and $b$ are perpendicular (orthogonal) if and only if:

$a \cdot b = 0$

This follows from the geometric definition $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ , since $cos 9 0^{\circ} = 0$ .

Solution

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

Step 1: Apply the perpendicularity condition $a \cdot b = 0$ .

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

Step 2: Expand and simplify.

$= 6 - 2 k - 4 = 2 - 2 k$

Step 3: Set equal to zero and solve for $k$ .

$2 - 2 k = 0 ⟹ 2 k = 2 ⟹ k = 1$

Answer: $k = 1$

Verification: With $k = 1$ : $a \cdot b = (2) (3) + (1) (- 2) + (- 1) (4) = 6 - 2 - 4 = 0 ✓$

The angle $θ$ between two vectors can be found using:

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

where $∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$ and $∣ b ∣ = b_{1}^{2} + b_{2}^{2} + b_{3}^{2}$ .

When $a \cdot b = 0$ , we get $cos θ = 0$ , confirming $θ = 9 0^{\circ}$ (perpendicular vectors).