$\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=p \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+q \hat{\mathbf{k}}$ and $\mathbf{d}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are given vectors. If…

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$\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=p \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+q \hat{\mathbf{k}}$ and $\mathbf{d}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are given vectors. If the projection of $\mathbf{c}$ on $\mathbf{a}$ is $5 \sqrt{3}$ units and if a, $\mathbf{b}$ and $\mathbf{c}$ form a parallelopiped of volume 5 cubic units, then $\tan ^{-1}(\mathbf{b}$. d $)=$

Aicharya · Accepted Answer

Given, $\begin{aligned} & \mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \ & \mathbf{c}=p \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+q \hat{\mathbf{k}}, \mathbf{d}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned}$ and projection of $\mathbf{c}$ on $\mathbf{a}=5 \sqrt{3}$ $\begin{aligned} & \frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{a}|}=5 \sqrt{3} \ & \Rightarrow \quad \frac{p+q-2}{\sqrt{3}}=5 \sqrt{3} \ & \Rightarrow \quad p+q-2=15 \ & \Rightarrow \quad p+q=17 \quad \ldots (i) \end{aligned}$ Also, given that $[\mathbf{a} \mathbf{b} \mathbf{c}]=5$ $\begin{array}{lrl} \Rightarrow & {\left[\begin{array}{ccc} 1 & -1 & 1 \ 1 & -2 & 1 \ p & 2 & q \end{array}\right]=5} \ \Rightarrow & 1(-2 q-2)+1(q-p)+1(2+2 p)=5 \ \Rightarrow & -2 q-2+q-p+2+2 p=5 \ \Rightarrow & p-q=5 \quad \ldots (ii) \end{array}$ By solving Eqs. (i) and (ii), we get $p=11, q=6$ Now, $\begin{aligned} \mathbf{b} \cdot \mathbf{d} & =(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \ & =p-2 q+2 \ & =11-12+2 \ & =1 \end{aligned}$ $\begin{aligned} \therefore \quad \tan ^{-1}(\mathbf{b} \cdot \mathbf{d}) & =\tan ^{-1}(\mathbf{l}) \ & =\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4} \end{aligned}$

\(\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-2…

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