Vector Algebra
Projection and Perpendicularity of Vectors
Grade 12
Question:
<p><strong>66.</strong> Let \(\vec{a} = \hat{i} + \hat{j} + \sqrt{2}\hat{k}\), \(\vec{b} = b_1\hat{i} + b_2\hat{j} + \sqrt{2}\hat{k}\) and \(\vec{c} = 5\hat{i} + \hat{j} + \sqrt{2}\hat{k}\) be three vectors such that projection vector of \(\vec{b}\) on \(\vec{a}\) is \(\vec{a}\). If \(\vec{a} + \vec{b}\) is perpendicular to \(\vec{c}\), then \(|\vec{b}|\) is equal to ________.</p>
Step-by-Step Solution
Key Concept: Use the projection condition: proj_a(b) = a implies b·a = |a|², and then apply the perpendicularity condition (a+b)·c = 0 to find the components of b, finally computing |b|.
Step 1: Find |a|^2 | a |^2 = 1^2 + 1^2 + (√2)^2 = 1 + 1 + 2 = 4 Step 2: Apply projection condition Given: proj_ a ( b ) = a This means: (( b · a )/| a |^2) a = a Therefore: b · a = | a |^2 = 4 Step 3: Calculate b·a in terms of unknowns b · a = b_1(1) + b_2(1) + √2(√2) = b_1 + b_2 + 2 So: b_1 + b_2 + 2 = 4 ⟹ b_1 + b_2 = 2 ... (i) Step 4: Apply perpendicularity condition ( a + b )· c = 0 ((1+b_1), (1+b_2), 2√2)·(5, 1, √2) = 0 5(1+b_1) + 1(1+b_2) + 2√2·√2 = 0 5 + 5b_1 + 1 + b_2 + 4 = 0 5b_1 + b_2 = -10 ... (ii) Step 5: Solve the system From (i): b_2 = 2 - b_1 Substitute in (ii): 5b_1 + (2 - b_1) = -10 4b_1 = -12 ⟹ b_1 = -3 b_2 = 2 - (-3) = 5 Step 6: Calculate |b| | b |^2 = (-3)^2 + 5^2 + (√2)^2 = 9 + 25 + 2 = 36 ∴ | b | = 6
Correct Answer: 6