The angle between the diagonals of the parallelogram whose adjacent sides are $2 \hat{i}+4 \hat{j}-5 \hat{k}…

The angle between the diagonals of the parallelogram whose adjacent sides are $2 \hat{i}+4 \hat{j}-5 \hat{k}, \hat{i}+2 \hat{j}+3 \hat{k}$ is
  1. $\cos ^{-1}\left(\frac{7}{\sqrt{69}}\right)$
  2. $\cos ^{-1}\left(\frac{1}{7 \sqrt{69}}\right)$.
  3. $\cos ^{-1}\left(\frac{1}{7}\right)$
  4. $\cos ^{-1}\left(\frac{31}{7 \sqrt{69}}\right)$

Solution

We know that diagonal of parallelograms are $\begin{aligned} & \vec{d}_1=\vec{a}+\vec{b}=3 \hat{i}+6 \hat{j}-2 \hat{k} \\ & \vec{d}_2=\vec{a}-\vec{b}=\hat{i}+2 \hat{j}-8 \hat{k} \end{aligned}$
Let angle between $\vec{d}_1$ and $\vec{d}_2$ is $\theta$ $\begin{aligned} & \therefore \cos \theta=\left|\frac{(3 \hat{i}+6 \hat{j}-2 \hat{k}) \cdot(\hat{i}+2 \hat{j}-8 \hat{k})}{\sqrt{9+36+4} \sqrt{1+4+64}}\right| \\ & =\frac{3+12+16}{\sqrt{49} \sqrt{69}}=\frac{31}{7 \sqrt{69}} \Rightarrow \theta=\cos ^{-1}\left(\frac{31}{7 \sqrt{69}}\right) \end{aligned}$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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