Vectors Question 85
Question: The vectors from origin to the points A and B are $ \overrightarrow{A}=3\hat{i}-6\hat{j}+2\hat{k} $ and $ \overrightarrow{B}=2\hat{i}+\hat{j}-2\hat{k} $ respectively. The area of the triangle OAB be
Options:
A) $ \frac{5}{2}\sqrt{17} $ sq.unit
B) $ \frac{2}{5}\sqrt{17} $ sq.unit
C) $ \frac{3}{5}\sqrt{17} $ sq.unit
D) $ \frac{5}{3}\sqrt{17} $ sq.unit
Correct Answer: A Given $ \overrightarrow{OA}=\overrightarrow{a}=3\hat{i}-6\hat{j}+2\hat{k} $ and $ \overrightarrow{OB}=\overrightarrow{b}=2\hat{i}+\hat{j}-2\hat{k} $ Therefore ,(\overrightarrow{a}\times \overrightarrow{b}),=| \begin{matrix} \hat{i} & \hat{j} & {\hat{k}} \\ ,3 & -6 & 2 \\ ,2, & 1 & -2, \\ \end{matrix} |, $ $ =(12-2)\hat{i}+(4+6)\hat{j}+(3+12)\hat{k} $ $ =10\hat{i}+10\hat{j}+15\hat{k} $ $ =\sqrt{425} $ $ =5\sqrt{17} $ Area of $ \Delta OAB=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|,=\frac{5\sqrt{17}}{2}, $ sq.unit.Show Answer
Answer:
Solution:
$ \
$ \Rightarrow |\overrightarrow{a}\times \overrightarrow{b}|,=,\sqrt{10^{2}+10^{2}+15^{2}} $
BETA
