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 $\alpha$…

Question

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

Aicharya · Accepted Answer

$ \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} $

Two adjacent sides of a parallelogram $A B C D$ are given by $\overrightarrow{\mathbf{A B}}=2…

Solution