Vector Algebra
Scalar Product of Vectors
Grade None
Question:
<p>If <span class="math">\(|\vec{a}| = 3\)</span>, <span class="math">\(|\vec{b}| = 4\)</span>, then find a value of <span class="math">\(\lambda\)</span> for which <span class="math">\(\vec{a} + \lambda \vec{b}\)</span> is perpendicular to <span class="math">\(\vec{a} - \lambda \vec{b}\)</span>.</p>
<p>(a) <span class="math">\(9/16\)</span></p>
<p>(b) <span class="math">\(3/4\)</span></p>
<p>(c) <span class="math">\(3/2\)</span></p>
<p>(d) <span class="math">\(4/3\)</span></p>
Step-by-Step Solution
Key Concept: Two vectors are perpendicular if and only if their scalar product equals zero. Use the distributive property of dot product to expand and simplify.
Solution: \(\vec{a} + \lambda \vec{b}\) is perpendicular to \(\vec{a} - \lambda \vec{b}\) . For perpendicularity: \((\vec{a} + \lambda \vec{b}) \cdot (\vec{a} - \lambda \vec{b}) = 0\) Expanding: \(\vec{a} \cdot \vec{a} - \lambda \vec{a} \cdot \vec{b} + \lambda \vec{b} \cdot \vec{a} - \lambda^2 \vec{b} \cdot \vec{b} = 0\) Since \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\) : \(|\vec{a}|^2 - \lambda^2 |\vec{b}|^2 = 0\) \(9 - 16\lambda^2 = 0\) \(\lambda^2 = \frac{9}{16}\) \(\lambda = \frac{3}{4}\) ∴ Answer is (b).
Correct Answer: B