A) \[2\,\mathbf{i}-2\,\mathbf{j}+\mathbf{k}\]
B) \[-\,2\,\mathbf{i}+2\,\mathbf{j}+\mathbf{k}\]
C) \[2\,\mathbf{i}+2\,\mathbf{j}-\mathbf{k}\]
D) None of these
Correct Answer: A
Solution :
Let the vector is \[x\mathbf{i}+y\mathbf{j}+z\mathbf{k}.\] Now according to the conditions, \[\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=3\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=9\] .....(i) \[6x+5y-2z=0\] .....(ii) and \[3x+y-4z=0\] .....(iii) \[[\because \]it is perpendicular to both vectors, hence by \[{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}=0]\] On solving the equation (i), (ii) and (iii), we get \[x=2,\] \[y=-2\] and \[z=1.\] Therefore, the required vector is \[2\mathbf{i}-2\mathbf{j}+\mathbf{k}.\] Trick : By inspection, the vector \[2\mathbf{i}-2\mathbf{j}+\mathbf{k}\] is of length 3 and also perpendicular to the given vectors.
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