JEE Advanced · Mathematics · 30. Vector Algebra
Two adjacent sides of a parallelogram \(A B C D\) are given by \(\overrightarrow{\mathbf{A B}}=2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{A D}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\). The side \(A D\) is rotated by an acute angle \(\alpha\) in the plane of the parallelogram so that \(A D\) becomes \(A D^{\prime}\). If \(A D^{\prime}\) makes a right angle with the side \(A B\), then the cosine of the angle \(\alpha\) is given by
- A
\(\frac{8}{9}\)
- B
\(\frac{\sqrt{17}}{9}\)
- C
\(\frac{1}{9}\)
- D
\(\frac{4 \sqrt{5}}{9}\)
Answer & Solution
Correct Answer
(B)
\(\frac{\sqrt{17}}{9}\)
Step-by-step Solution
Detailed explanation
\[
\overrightarrow{\mathbf{A B}}=2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}
\]

\[
\overrightarrow{\mathbf{A D}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}
\]
Angle ' \(\theta\) ' between \(\overrightarrow{\mathbf{A B}}\) and \(\overrightarrow{\mathbf{A D}}\) is
\[
\begin{aligned}
& \cos (\theta)=\left|\begin{array}{l}
\overrightarrow{\mathbf{A B}} \cdot \overrightarrow{\mathbf{A D}} \\
\mid \overrightarrow{\mathbf{A B}}\|\overrightarrow{\mathbf{A D}}\|
\end{array}\right| \\
& =\left|\frac{-2+20+22}{(15)(3)}\right|=\frac{8}{9} \\
& \Rightarrow \quad \sin (\theta)=\frac{\sqrt{17}}{9} \\
& \text { Since, } \quad \alpha+\theta=90^{\circ} \\
& \therefore \quad \cos (\alpha)=\cos \left(90^{\circ}-\theta\right) \\
& =\sin (\theta)=\frac{\sqrt{17}}{9} \\
&
\end{aligned}
\]
\overrightarrow{\mathbf{A B}}=2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}
\]

\[
\overrightarrow{\mathbf{A D}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}
\]
Angle ' \(\theta\) ' between \(\overrightarrow{\mathbf{A B}}\) and \(\overrightarrow{\mathbf{A D}}\) is
\[
\begin{aligned}
& \cos (\theta)=\left|\begin{array}{l}
\overrightarrow{\mathbf{A B}} \cdot \overrightarrow{\mathbf{A D}} \\
\mid \overrightarrow{\mathbf{A B}}\|\overrightarrow{\mathbf{A D}}\|
\end{array}\right| \\
& =\left|\frac{-2+20+22}{(15)(3)}\right|=\frac{8}{9} \\
& \Rightarrow \quad \sin (\theta)=\frac{\sqrt{17}}{9} \\
& \text { Since, } \quad \alpha+\theta=90^{\circ} \\
& \therefore \quad \cos (\alpha)=\cos \left(90^{\circ}-\theta\right) \\
& =\sin (\theta)=\frac{\sqrt{17}}{9} \\
&
\end{aligned}
\]
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