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

• A. \(\frac{8}{9}\)
• B. \(\frac{\sqrt{17}}{9}\)
• C. \(\frac{1}{9}\)
• D. \(\frac{4\sqrt{5}}{9}\)

Hint:

\(AD'\) lies in the plane and is perpendicular to \(AB\); find the angle between \(AD\) and \(AB\).

Answer 1

The Correct Option is B

Step 1: Resolve AD
Component of $\vec{AD}$ along $\vec{AB}$ has length $\dfrac{40}{15}=\dfrac83$. The perpendicular part has length $\sqrt{9-\tfrac{64}{9}}=\dfrac{\sqrt{17}}{3}$.

Step 2: Rotate
$AD'$ has length 3 and points along the perpendicular part, so $\cos\alpha=\dfrac{\sqrt{17}/3}{3}=\dfrac{\sqrt{17}}{9}$ (the dot product of $AD$ with $AD'$ divided by 9). Option (B).

Final Answer:
The cosine is $\sqrt{17}/9$, option (B). \[ \boxed{\dfrac{\sqrt{17}}{9}} \]