Search any question & find its solution
Question:
Answered & Verified by Expert
The vectors $\mathbf{A B}=3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}$ and $\mathbf{B C}=\mathbf{i}-2 \mathbf{k}$ are the adjacent sides of a parallelogram. The angle between its diagonals is
Options:
Solution:
2858 Upvotes
Verified Answer
The correct answer is:
None of these
Given, $\mathbf{A B}=3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}$
and $\quad \mathbf{B C}=\mathbf{i}-2 \mathbf{k}$
$\begin{aligned} \text { Diagonals } \mathbf{A C} & =\mathbf{A B}+\mathbf{B C} \\ & =3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}+\mathbf{i}-2 \mathbf{k} \\ & =4 \mathbf{i}-2 \mathbf{j}\end{aligned}$
and
$\begin{aligned} \mathbf{B D} & =\mathbf{B} \mathbf{C}-\mathbf{A} \mathbf{B} \\ & =\mathbf{i}-2 \mathbf{k}-(3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}) \\ & =-2 \mathbf{i}+2 \mathbf{j}-4 \mathbf{k}\end{aligned}$
$\therefore \quad \cos \theta=\frac{\mathbf{A C} \cdot \mathbf{B D}}{|\mathbf{A C}||\mathbf{B D}|}$
$\begin{aligned} & =\frac{(4 \mathbf{i}-2 \mathbf{j}) \cdot(-2 \mathbf{i}+2 \mathbf{j}-4 \mathbf{k})}{\sqrt{4^2+(-2)^2} \sqrt{(-2)^2+(2)^2+(-4)^2}} \\ & =\frac{-8-4}{\sqrt{20} \sqrt{24}}=-\frac{12}{4 \sqrt{30}}=-\frac{3}{\sqrt{30}}=-\sqrt{\frac{3}{10}} \\ & \Rightarrow \quad \theta=\cos ^{-1}\left(-\sqrt{\frac{3}{10}}\right) \\ & \text { or } \quad \pi-\cos ^{-1}\left(-\sqrt{\frac{3}{10}}\right) \\ & \end{aligned}$
and $\quad \mathbf{B C}=\mathbf{i}-2 \mathbf{k}$
$\begin{aligned} \text { Diagonals } \mathbf{A C} & =\mathbf{A B}+\mathbf{B C} \\ & =3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}+\mathbf{i}-2 \mathbf{k} \\ & =4 \mathbf{i}-2 \mathbf{j}\end{aligned}$
and
$\begin{aligned} \mathbf{B D} & =\mathbf{B} \mathbf{C}-\mathbf{A} \mathbf{B} \\ & =\mathbf{i}-2 \mathbf{k}-(3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}) \\ & =-2 \mathbf{i}+2 \mathbf{j}-4 \mathbf{k}\end{aligned}$
$\therefore \quad \cos \theta=\frac{\mathbf{A C} \cdot \mathbf{B D}}{|\mathbf{A C}||\mathbf{B D}|}$
$\begin{aligned} & =\frac{(4 \mathbf{i}-2 \mathbf{j}) \cdot(-2 \mathbf{i}+2 \mathbf{j}-4 \mathbf{k})}{\sqrt{4^2+(-2)^2} \sqrt{(-2)^2+(2)^2+(-4)^2}} \\ & =\frac{-8-4}{\sqrt{20} \sqrt{24}}=-\frac{12}{4 \sqrt{30}}=-\frac{3}{\sqrt{30}}=-\sqrt{\frac{3}{10}} \\ & \Rightarrow \quad \theta=\cos ^{-1}\left(-\sqrt{\frac{3}{10}}\right) \\ & \text { or } \quad \pi-\cos ^{-1}\left(-\sqrt{\frac{3}{10}}\right) \\ & \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.