Vector Algebra
Vector Magnitude and Coplanarity
Grade 12
Question:
<p><strong>Ex. 22</strong> If <strong>a</strong>, <strong>b</strong>, <strong>c</strong> be non-zero vectors such that <strong>a</strong> is perpendicular to <strong>b</strong> and <strong>c</strong> and \(|\mathbf{a}| = 1\), \(|\mathbf{b}| = 2\), \(|\mathbf{c}| = 1\), \(\mathbf{b} \cdot \mathbf{c} = 1\) and there is a non-zero vector <strong>d</strong> coplanar with \(\mathbf{a} + \mathbf{b}\) and \(2\mathbf{b} - \mathbf{c}\) and \(\mathbf{d} \cdot \mathbf{a} = 1\), then minimum value of \(|\mathbf{d}|\) is</p>
<p>(a) \(\frac{2}{\sqrt{13}}\)</p>
<p>(b) \(\frac{3}{\sqrt{13}}\)</p>
<p>(c) \(\frac{4}{\sqrt{13}}\)</p>
<p>(d) \(\frac{4\sqrt{13}}{13}\)</p>
Step-by-Step Solution
Key Concept: Express the vector d as a linear combination of the given coplanar vectors, apply the constraint d·a=1 to find one coefficient, and then minimize |d|² using calculus.
Given: \(\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} = 0\), \(|\mathbf{a}| = |\mathbf{c}| = 1\), \(|\mathbf{b}| = 2\) and \(\mathbf{b} \cdot \mathbf{c} = 1\) Step 1: Since d is coplanar with \(\mathbf{a} + \mathbf{b}\) and \(2\mathbf{b} - \mathbf{c}\), we can write: \(\mathbf{d} = x(\mathbf{a} + \mathbf{b}) + y(2\mathbf{b} - \mathbf{c})\) Step 2: Using the condition \(\mathbf{d} \cdot \mathbf{a} = 1\): \(x(1 + 0) + 0 = 1 \Rightarrow x = 1\) Step 3: Therefore, \(\mathbf{d} = \mathbf{a} + \mathbf{b} + y(2\mathbf{b} - \mathbf{c})\) Step 4: Calculate \(|\mathbf{d}|^2\): \[|\mathbf{d}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2\mathbf{a} \cdot \mathbf{b} + y^2(2\mathbf{b} - \mathbf{c}) \cdot (2\mathbf{b} - \mathbf{c}) + 2y(\mathbf{a} + \mathbf{b}) \cdot (2\mathbf{b} - \mathbf{c})\] \[= 1 + 4 + 0 + y^2(4|\mathbf{b}|^2 + |\mathbf{c}|^2 - 4\mathbf{b} \cdot \mathbf{c}) + 2y(2|\mathbf{b}|^2 - \mathbf{b} \cdot \mathbf{c})\] \[= 5 + y^2(16 + 1 - 4) + 2y(8 - 1) = 5 + 13y^2 - 14y\] Step 5: To minimize, take \(\frac{d}{dy}(5 + 13y^2 - 14y) = 0\): \(26y - 14 = 0 \Rightarrow y = \frac{7}{13}\) Step 6: Minimum value: \(|\mathbf{d}|_{\min}^2 = 5 + 13 \cdot \frac{49}{169} - 14 \cdot \frac{7}{13} = 5 + \frac{49}{13} - \frac{98}{13} = 5 - \frac{49}{13} = \frac{16}{13}\) \(\therefore |\mathbf{d}|_{\min} = \frac{4}{\sqrt{13}} = \frac{4\sqrt{13}}{13}\)
Correct Answer: d