Kinematics Question 436
Question: The vectors from origin to the points A and B are $ \vec{A}=3\hat{i}-6\hat{j}+2\hat{k} $ and $ \vec{B}=2\hat{i}+\hat{j}-2\hat{k} $ respectively. The area of the triangle OAB be
Options:
A) $ \frac{5}{2}\sqrt{17}squnit $
B) $ \frac{2}{5}\sqrt{17}squnit $
C) $ \frac{3}{5}\sqrt{17}squnit $
D) $ \frac{5}{3}\sqrt{17}squnit $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given $ \overrightarrow{OA}=\vec{a}=3\hat{i}-6\hat{j}+2\hat{k} $
and $ \overrightarrow{OB}=\vec{b}=2\hat{i}+\hat{j}-2\hat{k} $
$ \therefore $ $ (\vec{a}\times \vec{b})= \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & -6 & 2 \\ 2 & 1 & -2 \\ \end{vmatrix} $ $ =(12-2)\hat{i}+(4+6)\hat{j}+(3+12)\hat{k} $ $ =10\hat{i}+10\hat{j}+15\hat{k} $
$ \Rightarrow $$ |\vec{a}\times \vec{b}|=\sqrt{10^{2}+10^{2}+15^{2}} $
$ =\sqrt{425}=5\sqrt{17} $
Area of $ \Delta OAB=\frac{1}{2}|\vec{a}\times \vec{b}|=\frac{5\sqrt{17}}{2}sq.\ unit $
BETA
