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  • The vectors from origin to the points A and B are a = 2i - 3j + 2k and b = 2i+ 3j + k respectively, then the area of triangle OAB is A. 340

The vectors from origin to the points A and B are \vec{\mathrm{a}}=2 \hat{\mathrm{i}}-3\hat{\mathrm{j}}+2\hat{\mathrm{k}} \text { and }\vec{\mathrm{b}}=2\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}} respectively, then the area of triangle OAB is
A. 340
B. \sqrt{25}
C. \sqrt{229}
D. \frac{1}{2}\sqrt{229}

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Answer :(D)

Given that, vector from origin to the point A, \vec{OA}=2 \hat{\mathrm{i}}-3\hat{\mathrm{j}}+2\hat{\mathrm{k}} and vector from origin to the point B,  \vec{OB}=2\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}

\\ \text { Area of } \Delta \mathrm{OAB}=\frac{1}{2}|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}| \\ =\frac{1}{2}|(2 \hat{\mathrm{l}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})|

\\ =\frac{1}{2}\left|\begin{array}{ccc} \hat{1} & \hat{\jmath} & \hat{\mathrm{k}} \\ 2 & -3 & 2 \\ 2 & 3 & 1 \end{array}\right| \\ =\frac{1}{2}|\hat{\imath}(-3-6)-\hat{\jmath}(2-4)+\hat{\mathrm{k}}(6+6)| \\ =\frac{1}{2}|(-9 \hat{\imath}+2 \hat{\jmath}+12 \hat{\mathrm{k}})| \\ =\frac{1}{2} \sqrt{(-9)^{2}+(2)^{2}+(12)^{2}}

\\ =\frac{1}{2} \sqrt{81+4+144} \\ =\frac{1}{2} \sqrt{229}

 

 

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