Vector Algebra
Vector Projections
Grade 12
Question:
<p><strong>Ex. 97</strong> Let <strong>a</strong> = 2<strong>i</strong> + 3<strong>j</strong> - 6<strong>k</strong>, <strong>b</strong> = 2<strong>i</strong> - 3<strong>j</strong> + 6<strong>k</strong> and <strong>c</strong> = -2<strong>i</strong> + 3<strong>j</strong> + 6<strong>k</strong>. Let <strong>a</strong>₁ be the projection of <strong>a</strong> on <strong>b</strong> and <strong>a</strong>₂ be the projection of <strong>a</strong>₁ on <strong>c</strong>. Then <strong>a</strong>₂ is equal to</p>
<p>(a) \(\frac{943}{49}(2\mathbf{i} - 3\mathbf{j} - 6\mathbf{k})\)</p>
<p>(b) \(\frac{943}{49} \cdot 2(2\mathbf{i} - 3\mathbf{j} - 6\mathbf{k})\)</p>
<p>(c) \(\frac{943}{49}(-2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k})\)</p>
<p>(d) \(\frac{943}{49}(-2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k})\)</p>
Step-by-Step Solution
Key Concept: The projection of a vector onto another is found using the formula: projection = (dot product / magnitude squared) times the direction vector. Apply this twice in sequence.
Step 1: Calculate the projection of a on b : \(\mathbf{a}_1 = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b}\) Step 2: \(\mathbf{a} \cdot \mathbf{b} = (2)(2) + (3)(-3) + (-6)(6) = 4 - 9 - 36 = -41\) Step 3: \(|\mathbf{b}|^2 = 4 + 9 + 36 = 49\) Step 4: \(\mathbf{a}_1 = \frac{-41}{49}(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k})\) Step 5: Calculate the projection of a _1 on c : \(\mathbf{a}_2 = \frac{\mathbf{a}_1 \cdot \mathbf{c}}{|\mathbf{c}|^2} \mathbf{c}\) Step 6: \(\mathbf{a}_1 \cdot \mathbf{c} = \frac{-41}{49}[(2)(-2) + (-3)(3) + (6)(6)] = \frac{-41}{49}(-4 - 9 + 36) = \frac{-41 \cdot 23}{49}\) Step 7: \(|\mathbf{c}|^2 = 4 + 9 + 36 = 49\) Step 8: \(\mathbf{a}_2 = \frac{-41 \cdot 23}{49 \cdot 49}(-2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) = \frac{943}{49}(-2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k})\) ∴ Answer is (c).
Correct Answer: C