Search any question & find its solution
Question:
Answered & Verified by Expert
In a triangle $A B C$, the sides $A B$ and $A C$ are represented by the vectors $3 \mathrm{i}+\mathrm{j}+\mathrm{k}$ and $\mathrm{i}+2 \mathrm{j}+\mathrm{k}$, respectively. Calculate the angle $\angle A B C$
Options:
Solution:
1574 Upvotes
Verified Answer
The correct answer is:
$\cos ^{-1} \sqrt{\frac{5}{11}}$
$A B=3 i+j+\hat{k}$ and $A C=i+2 j+i$
$8 A=-(3 i+j+k)$
$\angle A B C$ is angle between $B A$ and $B C$
$B C=A C+B A$
$=A C-A B$
$=\hat{i}+2 \hat{j}+\hat{k}-(3 \hat{i}+\hat{j}+\hat{k})=-2 \hat{i}+\hat{j}$
$B A \cdot B C=|B A \| B C| \cos \theta$
$-(3 \hat{i}+\hat{j}+\hat{k}) \cdot(-2 \hat{i}+\hat{j})$
$=\left(\sqrt{3^{2}+1^{2}+1^{2}}\right) \cdot\left(\sqrt{2^{2}+1^{2}}\right) \cos \theta$
$\Rightarrow \quad 6-1=\sqrt{11 \times 5} \cdot \cos \theta$
$\cos \theta=\frac{5}{\sqrt{55}}=\sqrt{\frac{5}{11}}=\theta=\cos ^{-1}=\sqrt{\frac{5}{11}}$
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.