A) \[-1\]
B) \[0\]
C) \[1\]
D) None of these
Correct Answer: D
Solution :
\[\overrightarrow{AB}=2\hat{i}+\hat{j}+\hat{k},\]\[\overrightarrow{AC}=(t+1)\,\hat{i}+0\hat{j}+\hat{k}\] \[\overrightarrow{AB}\times \overrightarrow{AC}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & 1 & 1 \\ t+1 & 0 & -1 \\ \end{matrix} \right|=-\hat{i}+(t+3)\hat{j}-(t+1)\hat{k}\] \[|\overrightarrow{AB}\times \overrightarrow{AC}|\,=\,\sqrt{1+{{(t+3)}^{2}}+{{(t+1)}^{2}}}\] \[=\sqrt{2{{t}^{2}}+8t+11}\] Area of\[\Delta \,ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|\] \[\Rightarrow \] \[\Delta =\frac{1}{2}\sqrt{2{{t}^{2}}+8t+11}\] Let \[f(t)={{\Delta }^{2}}=\frac{1}{4}(2{{t}^{2}}+8t+11)\] \[f'(t)=0\Rightarrow t=-\,2\] at \[t=-\,2,\,f''(t)>0\] So \[\Delta \]is min at\[t=-\,2\Rightarrow \]max, area\[=\sqrt{3}/2\]
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