Search any question & find its solution
Question:
Answered & Verified by Expert
If $|\bar{a}|=4,|\bar{b}|=5,|\bar{a}-\bar{b}|=3$ and $\theta$ is the angle between the vectors $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$, then $\cot ^2 \theta=$
Options:
Solution:
1916 Upvotes
Verified Answer
The correct answer is:
$\frac{9}{16}$
$|\vec{a}|=4,|\vec{b}|=5$
$\begin{aligned} & |\vec{a}-\vec{b}|=3 \Rightarrow|\vec{a}-\vec{b}|^2=9 \\ & (\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})=9 \\ & |\vec{a}|^2-2|\vec{a} \| \vec{b}| \cos \theta+|\vec{b}|^2=9 \\ & 16+25-2(4)(5) \cos \theta=9 \\ & \cos \theta=\frac{4}{5} \Rightarrow \cot \theta=\frac{4}{3} ; \cot ^2 \theta=\frac{16}{9}\end{aligned}$
$\begin{aligned} & |\vec{a}-\vec{b}|=3 \Rightarrow|\vec{a}-\vec{b}|^2=9 \\ & (\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})=9 \\ & |\vec{a}|^2-2|\vec{a} \| \vec{b}| \cos \theta+|\vec{b}|^2=9 \\ & 16+25-2(4)(5) \cos \theta=9 \\ & \cos \theta=\frac{4}{5} \Rightarrow \cot \theta=\frac{4}{3} ; \cot ^2 \theta=\frac{16}{9}\end{aligned}$
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.