A) \[5\mathbf{i}+5\mathbf{j}+\mathbf{k}\]
B) \[5\mathbf{i}+\mathbf{j}-5\mathbf{k}\]
C) \[5\mathbf{i}+\mathbf{j}+5\mathbf{k}\]
D) \[\pm \,(5\mathbf{i}-\mathbf{j}-5\mathbf{k})\]
Correct Answer: D
Solution :
Let the required vector be \[\alpha ={{d}_{1}}\mathbf{i}+{{d}_{2}}\mathbf{j}+{{d}_{3}}\mathbf{k}\], where \[d_{1}^{2}+d_{2}^{2}+d_{3}^{2}=51\], (given) .....(i) Now, each of the given vectors \[\mathbf{a},\,\,\mathbf{b},\,\,\mathbf{c}\] is a unit vector \[\cos \theta =\frac{\mathbf{d}\,.\,\mathbf{a}}{|\mathbf{d}|\,|\mathbf{a}|}=\frac{\mathbf{d}\,.\,\mathbf{b}}{|\mathbf{d}|\,|\mathbf{b}|}=\frac{\mathbf{d}\,.\,\mathbf{c}}{|\mathbf{d}|\,|\mathbf{c}|}\] or \[\mathbf{d}\,.\,\mathbf{a}=\mathbf{d}\,.\,\mathbf{b}=\mathbf{d}\,.\,\mathbf{c}\] \[|\mathbf{d}|=\sqrt{51}\] cancels out and \[|\mathbf{a}|\,\,=\,\,|\mathbf{b}|\,=\,|\mathbf{c}|\,=1\] Hence, \[\frac{1}{3}({{d}_{1}}-2{{d}_{2}}+2{{d}_{3}})=\frac{1}{5}(-4{{d}_{1}}+0{{d}_{2}}-3{{d}_{3}})={{d}_{2}}\] \[\Rightarrow {{d}_{1}}-5{{d}_{2}}+2{{d}_{3}}=0\] and \[4{{d}_{1}}+5{{d}_{2}}+3{{d}_{3}}=0\] On solving, \[\frac{{{d}_{1}}}{5}=\frac{{{d}_{2}}}{-1}=\frac{{{d}_{3}}}{-5}=\lambda \] (say) Putting \[{{d}_{1}},\,{{d}_{2}}\] and \[{{d}_{3}}\] in (i), we get \[\lambda =\pm 1\] Hence the required vectors are \[\pm (5\mathbf{i}-\mathbf{j}-5\mathbf{k}).\] Trick: Check it with the options.
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