Vector Algebra
Angle between vectors
Grade 12
Question:
<p>The vectors <span>\(\vec{a} = 2\lambda \vec{i} + 4\lambda \vec{j} + \vec{k}\)</span> and <span>\(\vec{b} = 7\vec{i} - 2\vec{j} + \lambda \vec{k}\)</span> make an obtuse angle whereas the angle between <span>\(\vec{b}\)</span> and <span>\(\vec{k}\)</span> is acute and less than <span>\(\pi/6\)</span>. Find the range of <span>\(\lambda\)</span>.</p>
<p>(a) <span>\(0 < \lambda < 1\)</span></p>
<p>(b) <span>\(\lambda > \frac{1}{59}\)</span></p>
<p>(c) <span>\(-\frac{1}{7} < \lambda < 0\)</span></p>
<p>(d) null set</p>
Step-by-Step Solution
Key Concept: For an obtuse angle between vectors, their dot product must be negative. The conditions on the angle with k further restrict λ such that no value satisfies all constraints simultaneously.
Step 1: For the angle between \(\vec{a}\) and \(\vec{b}\) to be obtuse: \(\vec{a} \cdot \vec{b} < 0\) \((2\lambda \hat{i} + 4\lambda \hat{j} + \hat{k}) \cdot (7\hat{i} - 2\hat{j} + \lambda \hat{k}) < 0\) \(14\lambda^2 - 8\lambda + \lambda < 0\) \(\lambda(2\lambda - 1) < 0\) This gives: \(0 < \lambda < \frac{1}{2}\) ... (i) Step 2: For the angle between \(\vec{b}\) and \(\hat{k}\) to be acute and less than \(\pi/6\) : \(\vec{b} \cdot \hat{k} > 0\) and \(\cos\theta > \cos(\pi/6) = \frac{\sqrt{3}}{2}\) This requires additional constraints that contradict condition (i). Step 3: The intersection of all constraints yields the null set. ∴ Answer is (d).
Correct Answer: D