A) \[13\,\,sq\,\,unit\]
B) \[\sqrt{13}\,\,sq\,\,unit\]
C) \[56\,\,sq\,\,unit\]
D) \[\sqrt{6}\,\,sq\,\,unit\]
Correct Answer: B
Solution :
If \[O\] be the point of origin, the position vectors of a vertices are \[\overset{\to }{\mathop{\mathbf{OA}}}\,=\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+2\widehat{\mathbf{k}},\,\,\overset{\to }{\mathop{\mathbf{OB}}}\,=2\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-\widehat{\mathbf{k}}\] \[\overset{\to }{\mathop{\mathbf{OC}}}\,=3\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+2\widehat{\mathbf{k}}\] Now, \[\overset{\to }{\mathop{\mathbf{AB}}}\,=2\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-\widehat{\mathbf{k}}-(-\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+2\widehat{\mathbf{k}})\] \[=\widehat{\mathbf{i}}+2\widehat{\mathbf{j}}-3\widehat{\mathbf{k}}\] and \[\overset{\to }{\mathop{\mathbf{AC}}}\,=3\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+2\widehat{\mathbf{k}}-(\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+2\widehat{\mathbf{k}})\] \[=2\widehat{\mathbf{i}}\] \[\therefore \]Area of triangle\[=\frac{1}{2}||\overset{\to }{\mathop{\mathbf{AB}}}\,\times \overset{\to }{\mathop{\mathbf{AC}}}\,||\] \[=\frac{1}{2}\left| \left| \begin{matrix} \widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \\ 1 & 2 & -3 \\ 2 & 0 & 0 \\ \end{matrix} \right| \right|\] \[=\frac{1}{2}|[\widehat{\mathbf{i}}(0)-\widehat{\mathbf{j}}(6)+\widehat{\mathbf{k}}(-4)]|\] \[=\frac{1}{2}|[-6\widehat{\mathbf{i}}-4\widehat{\mathbf{k}}]|\] \[=\frac{1}{2}\sqrt{36+16}=\frac{1}{2}\times 2\sqrt{13}\] \[=\sqrt{13}sq\,\,unit\] Note: If area of triangle is zero, then points are collinear.You need to login to perform this action.
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