A) \[3\,\,sq\,\,unit\]
B) \[3\sqrt{3}sq\,\,unit\]
C) \[4\,\,sq\,\,unit\]
D) \[5\,\,sq\,\,unit\]
Correct Answer: A
Solution :
Key Idea: (i) The area of a triangle with adjacent sides \[\overset{\to }{\mathop{\mathbf{a}}}\,\] and \[\overset{\to }{\mathop{\mathbf{b}}}\,\] is\[\frac{1}{2}|\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{b}}}\,|\]. (ii) The area of a triangle \[ABC\] is\[\frac{1}{2}|\overset{\to }{\mathop{\mathbf{A}}}\,\overset{\to }{\mathop{\mathbf{B}}}\,\times \overset{\to }{\mathop{\mathbf{A}}}\,\overset{\to }{\mathop{\mathbf{C}}}\,|\] or \[\frac{1}{2}|\overset{\to }{\mathop{\mathbf{B}}}\,\overset{\to }{\mathop{\mathbf{C}}}\,\times \overset{\to }{\mathop{\mathbf{A}}}\,\overset{\to }{\mathop{\mathbf{B}}}\,|\]or\[\frac{1}{2}|\overset{\to }{\mathop{\mathbf{C}}}\,\overset{\to }{\mathop{\mathbf{B}}}\,\times \overset{\to }{\mathop{\mathbf{B}}}\,\overset{\to }{\mathop{\mathbf{A}}}\,|\] Let \[\overset{\to }{\mathop{\mathbf{OA}}}\,=\widehat{\mathbf{i}},\,\,\overset{\to }{\mathop{\mathbf{OB}}}\,=\widehat{\mathbf{i}}+2\mathbf{\hat{j}},\,\,\overset{\to }{\mathop{\mathbf{OC}}}\,=\mathbf{\hat{i}}+3\mathbf{\hat{k}}\] \[\therefore \] \[\overset{\to }{\mathop{\mathbf{AB}}}\,=\overset{\to }{\mathop{\mathbf{OB}}}\,-\overset{\to }{\mathop{\mathbf{OA}}}\,=(\widehat{\mathbf{i}}+2\widehat{\mathbf{j}})-(\widehat{\mathbf{i}})=2\widehat{\mathbf{j}}\] and \[\overset{\to }{\mathop{\mathbf{AC}}}\,=\overset{\to }{\mathop{\mathbf{OC}}}\,-\overset{\to }{\mathop{\mathbf{OA}}}\,=(\widehat{\mathbf{i}}+3\widehat{\mathbf{k}})-(\widehat{\mathbf{i}})=3\widehat{\mathbf{k}}\] \[\therefore \]Area of\[\Delta ABC=\frac{1}{2}|\overset{\to }{\mathop{\mathbf{A}}}\,\overset{\to }{\mathop{\mathbf{B}}}\,\times \overset{\to }{\mathop{\mathbf{A}}}\,\overset{\to }{\mathop{\mathbf{C}}}\,|\] \[=\frac{1}{2}\left| \begin{matrix} \widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{matrix} \right|=\frac{1}{2}[\widehat{\mathbf{i}}(6)-\widehat{\mathbf{j}}(0)+\widehat{\mathbf{k}}(0)]\] \[=\frac{1}{2}\sqrt{36+0+0}=\frac{6}{2}=3\,\,sq\,\,unit\]
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