Vector Algebra
Dot Product and Orthogonality
Grade 12
Question:
<p>If \(a\), \(b\) and \(c\) are \(p\)th, \(q\)th, \(r\)th terms of HP and \(\vec{u} = (q - r)\vec{i} + (r - p)\vec{j} + (p - q)\vec{k}\), \(\vec{v} = \frac{1}{a}\vec{i} + \frac{1}{b}\vec{j} + \frac{1}{c}\vec{k}\), then</p>
<p>(a) \(\vec{u}\) and \(\vec{v}\) are parallel vectors</p>
<p>(b) \(\vec{u}\) and \(\vec{v}\) are orthogonal vectors</p>
<p>(c) \(\vec{u} \cdot \vec{v} = 1\)</p>
<p>(d) \(\vec{u} \times \vec{v} = \vec{i} + \vec{j} + \vec{k}\)</p>
Step-by-Step Solution
Key Concept: Recognize that if terms are in HP, their reciprocals are in AP, which creates a dot product relationship that equals zero.
Step 1: Since \(a\), \(b\), \(c\) are in HP, their reciprocals \(\frac{1}{a}\), \(\frac{1}{b}\), \(\frac{1}{c}\) are in AP. Step 2: If they are \(p\)th, \(q\)th, \(r\)th terms of HP, then \(\frac{1}{a}\), \(\frac{1}{b}\), \(\frac{1}{c}\) are \(p\)th, \(q\)th, \(r\)th terms of an AP. Step 3: For an AP: \(\frac{1}{b} - \frac{1}{a} = (q-p)d\) and \(\frac{1}{c} - \frac{1}{b} = (r-q)d\) for common difference \(d\). Step 4: Compute \(\vec{u} \cdot \vec{v} = (q-r)\frac{1}{a} + (r-p)\frac{1}{b} + (p-q)\frac{1}{c} = 0\) (by the AP property). ∴ \(\vec{u}\) and \(\vec{v}\) are orthogonal vectors. Answer is (b).
Correct Answer: B