Question

Two adjacent sides of a parallelogram \(A B C D\) are given by \(\overline{\mathrm{AB}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}\) and \(\overline{\mathrm{AD}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\). The side AD is rotated by an acute angle \(\alpha\) in the plane of parallelogram so that AD becomes \(\mathrm{AD}^{\prime}\). If \(\mathrm{AD}^{\prime}\) makes a right angle with the side AB , then \(\cos \alpha=\)

• A. \(\frac{\sqrt{17}}{8}\)
• B. \(\frac{\sqrt{17}}{9}\)
• C. \(\frac{\sqrt{17}}{13}\)
• D. \(\frac{\sqrt{17}}{16}\)

Hint:

Find the angle between AB and AD from the dot product; the rotation needed to reach perpendicular is 90 degrees minus that angle.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Let \(\theta\) be the angle between \(\overline{AB}\) and \(\overline{AD}\). Rotating AD towards AB until it is perpendicular to AB needs an acute angle \(\alpha=90^{\circ}-\theta\).

Step 2: Find cos theta
\(\overline{AB}\cdot\overline{AD}=-2+20+22=40\), \(|\overline{AB}|=\sqrt{4+100+121}=15\), \(|\overline{AD}|=\sqrt{1+4+4}=3\).
\(\cos\theta=\frac{40}{45}=\frac{8}{9}\).

Step 3: Find cos alpha
\(\cos\alpha=\cos(90^{\circ}-\theta)=\sin\theta=\sqrt{1-\frac{64}{81}}=\frac{\sqrt{17}}{9}\).

Step 4: Compare the options
This is option (B).

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