Find the angle between the diagonals of parallelogram $P Q R S$, if $\mathbf{P Q}=3 \hat{\mathbf{i}}-2…

1. Only $\cos \theta=-\sqrt{\frac{3}{10}}$
2. Both $\cos \theta= \pm \sqrt{\frac{3}{10}}$
3. $\tan \theta=-\sqrt{\frac{3}{10}}$
4. $\tan \theta=-\sqrt{\frac{11}{10}}$

Solution

$ \text { 60. (a) } \begin{aligned} \mathbf{P Q} & =3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \mathbf{P S} & =\hat{\mathbf{i}}-2 \hat{\mathbf{k}} \end{aligned} $
$ \begin{aligned} & \text { Eqs. (i) }+ \text { (ii) } \Rightarrow \mathbf{P Q}+\mathbf{P S}=4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\mathbf{0} \hat{\mathbf{k}} \\ & \mathbf{P R}=4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}} \\ & \text { Eqs. (i) }- \text { (ii) } \Rightarrow \mathbf{P Q}-\mathbf{P S}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ & -\mathbf{Q S}=\mathbf{2} \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ & \Rightarrow \quad \mathbf{Q S}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ & \end{aligned} $ Let $\boldsymbol{\theta}$ be the angle between diagonal $\mathbf{P R}$ and $\mathbf{Q S}$ $ \begin{aligned} \therefore \quad \cos \theta & =\frac{\mathbf{P R} \cdot \mathbf{Q S}}{|\mathbf{P R}||\mathbf{Q S}|} \\ & =\frac{(4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}})(-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})}{\sqrt{16+4} \sqrt{4+4+16}} \\ & =\frac{-8-4}{\sqrt{20} \cdot \sqrt{24}}=\frac{-12}{2 \sqrt{5} \cdot 2 \sqrt{6}} \\ & =\frac{-3}{\sqrt{30}}=-\sqrt{\frac{9}{30}}=-\sqrt{\frac{3}{10}} \\ \therefore \quad \cos \theta & =-\sqrt{\frac{3}{10}} \end{aligned} $ Hence, option (1) is correct

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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