A) 7
B) 14
C) \[\frac{7}{2}\]
D) 4
Correct Answer: C
Solution :
Clearly minimum value of \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]. \[={{\left( \frac{|3(0)+2(0)+(0)-7|}{\sqrt{{{(3)}^{2}}\,+{{(2)}^{2}}+{{(1)}^{2}}}} \right)}^{2}}=\frac{49}{14}\,=\frac{7}{2}\] units. (This is possible when P(a, b, c) is foot of perpendicular from O(0, 0, 0) on the plane.) Alternatively: Let \[{{\vec{V}}_{1}}=3\hat{i}+2\hat{j}+\hat{k}\] and \[{{\vec{V}}_{2}}=a\hat{i}+b\hat{j}+c\hat{k}\] Now \[{{\vec{V}}_{1}}.{{\vec{V}}_{2}}=3a+2b+c\,=7\le |{{\vec{V}}_{1}}||{{\vec{V}}_{2}}|\] \[\Rightarrow \,\,7\le \,\sqrt{14}\,\times \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\] \[\Rightarrow \,({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\ge \frac{49}{14}\] Hence, \[{{\left. {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right|}_{least}}\,=\frac{7}{2}\]
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