Question:

Two adjacent sides of a parallelogram ABCD are given by \(\overline{AB} = 2\hat{i}+10\hat{j}+11\hat{k}\) and \(\overline{AD} = -\hat{i}+2\hat{j}+2\hat{k}\). The side AD is rotated by an acute angle \(α\) in the plane of the parallelogram so that AD becomes AD'. If AD' makes a right angle with the side AB, then the cosine of the angle \(α\) is given by

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\(AD'\) lies in the plane and is perpendicular to \(AB\); find the angle between \(AD\) and \(AB\).
Updated On: Oct 1, 2026
  • \(\frac{8}{9}\)
  • \(\frac{\sqrt{17}}{9}\)
  • \(\frac{1}{9}\)
  • \(\frac{4\sqrt{5}}{9}\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept
Let \(\varphi\) be the angle between \(\vec{AB}\) and \(\vec{AD}\). The new side \(AD'\) is in the same plane, perpendicular to \(AB\), and obtained by an acute rotation.

Step 2: Key Formula or Approach
\[ \cos\varphi=\frac{\vec{AB}\cdot\vec{AD}}{|\vec{AB}||\vec{AD}|} \]

Step 3: Detailed Explanation
\(\vec{AB}\cdot\vec{AD}=-2+20+22=40\), \(|\vec{AB}|=\sqrt{4+100+121}=15\), \(|\vec{AD}|=3\).
\(\cos\varphi=\dfrac{40}{45}=\dfrac89\), so \(\sin\varphi=\dfrac{\sqrt{17}}{9}\).
To reach a direction at \(90^{\circ}\) from \(AB\), the acute rotation is \(\alpha=90^{\circ}-\varphi\).
\[ \cos\alpha=\sin\varphi=\frac{\sqrt{17}}{9} \]

Final Answer:
\(\cos\alpha=\dfrac{\sqrt{17}}{9}\), option (B). \[ \boxed{\dfrac{\sqrt{17}}{9}\ \text{(B)}} \]
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