Vector Algebra
Projection and perpendicularity of vectors
Grade 12
Question:
<p>Given <br/> \(\vec{a} = \hat{i} + \hat{j} + \sqrt{2}\hat{k}\), \(\vec{b} = b_1\hat{i} + b_2\hat{j} + \sqrt{2}\hat{k}\) and \(\vec{c} = 5\hat{i} + \hat{j} + \sqrt{2}\hat{k}\).<br/> Projection of \(\vec{b}\) on \(\vec{a}\) is \(|\vec{a}|\) and \((\vec{a}+\vec{b})\) is perpendicular to \(\vec{c}\). Find \(|\vec{b}|\).</p>
Step-by-Step Solution
Key Concept: Use the projection formula constraint to establish a relationship between components of b, then apply the perpendicularity condition to get another equation. These two constraints simultaneously determine b₁ and b₂.
Step 1: Find |a| |a| = √(1^2 + 1^2 + (√2)^2) = √(1 + 1 + 2) = √4 = 2 Step 2: Use projection condition Projection of b on a = (b·a)/|a| = |a| = 2 So: b·a = 2|a| = 2(2) = 4 b·a = b_1(1) + b_2(1) + √2(√2) = b_1 + b_2 + 2 = 4 Therefore: b_1 + b_2 = 2 ... (1) Step 3: Use perpendicularity condition (a + b) ⊥ c means (a + b)·c = 0 a + b = (1 + b_1)i + (1 + b_2)j + 2√2k (a + b)·c = (1 + b_1)(5) + (1 + b_2)(1) + 2√2(√2) = 0 5 + 5b_1 + 1 + b_2 + 4 = 0 5b_1 + b_2 = -10 Therefore: 5b_1 + b_2 = -10 ... (2) Step 4: Solve the system From (2) - (1): 4b_1 = -12, so b_1 = -3 From (1): b_2 = 2 - (-3) = 5 Step 5: Find |b| |b| = √((-3)^2 + 5^2 + (√2)^2) = √(9 + 25 + 2) = √36 = 6
Correct Answer: 6