Vector Algebra
Direction Cosines and Unit Vectors
Grade 12
Question:
<p><strong>Example 33.</strong> The vector \(\mathbf{c}\), directed along the internal bisector of the angle between the vectors \(\mathbf{a} = 7\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}\) and \(\mathbf{b} = -2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) with \(|\mathbf{c}| = 5\sqrt{6}\), is</p>
<p>(a) \(\frac{5}{3}(\mathbf{i} - 7\mathbf{j} + 2\mathbf{k})\)</p>
<p>(b) \(\frac{5}{3}(5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k})\)</p>
<p>(c) \(\frac{5}{3}(\mathbf{i} + 7\mathbf{j} + 2\mathbf{k})\)</p>
<p>(d) \(\frac{5}{3}(-5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k})\)</p>
Step-by-Step Solution
Key Concept: The direction of the angle bisector between two vectors is proportional to the sum of their unit vectors. Use this property and the given magnitude to find the required vector.
Step 1: The vector along the internal bisector of angle between \(\mathbf{a}\) and \(\mathbf{b}\) is given by: \(\mathbf{c} = \lambda\left(\frac{\mathbf{a}}{|\mathbf{a}|} + \frac{\mathbf{b}}{|\mathbf{b}|}\right)\) Step 2: Calculate magnitudes: \(|\mathbf{a}| = \sqrt{49 + 16 + 16} = \sqrt{81} = 9\) \(|\mathbf{b}| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\) Step 3: The unit vectors are: \(\frac{\mathbf{a}}{|\mathbf{a}|} = \frac{1}{9}(7\mathbf{i} - 4\mathbf{j} - 4\mathbf{k})\) \(\frac{\mathbf{b}}{|\mathbf{b}|} = \frac{1}{3}(-2\mathbf{i} - \mathbf{j} + 2\mathbf{k})\) Step 4: The direction along bisector is proportional to \(\mathbf{i} - 7\mathbf{j} + 2\mathbf{k}\) Step 5: Using \(|\mathbf{c}| = 5\sqrt{6}\), we get \(\mathbf{c} = \frac{5}{3}(\mathbf{i} - 7\mathbf{j} + 2\mathbf{k})\) ∴ Answer is A.
Correct Answer: A