A) i ? k
B) i ? j ? 2k
C) i + j ? k
D) 3i + 3j ? 6k
Correct Answer: B
Solution :
Let vector be \[a\mathbf{i}+b\mathbf{j}+c\mathbf{k}\]. \[\because a\mathbf{i}+b\mathbf{j}+c\mathbf{k}\,,\,\mathbf{i}+\mathbf{j}\,,\,\,\mathbf{j}+\mathbf{k}\] are coplanar. \ \[\left| \,\begin{matrix} a & b & c \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{matrix}\, \right|=0\] \[\Rightarrow a-b+c=0\] Also, since \[(a\mathbf{i}+b\mathbf{j}+c\mathbf{k}\,)|\,|\,\,(2\mathbf{i}-2\mathbf{j}-4\mathbf{k})\] \ \[(a\mathbf{i}+b\mathbf{j}+c\mathbf{k})\times (2\mathbf{i}-2\mathbf{j}-4\mathbf{k})=0\] i.e., \[\left| \,\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & b & c \\ 2 & -2\, & -\,4\,\, \\ \end{matrix}\, \right|=0\] Þ \[\mathbf{i}(-\,4b+2c)-\mathbf{j}(-\,4a-2c)+\mathbf{k}(-\,2a-2b)=0\] Þ \[-\,4b+2c=0,\,\,4a+2c=0,\,\,\,2a+2b=0\] Þ \[\frac{c}{2}=\frac{b}{1},\] \[\frac{c}{2}=\frac{a}{-1},\] \[\frac{a}{-1}=\frac{b}{1}\] i.e., \[\frac{a}{-1}=\frac{b}{1}=\frac{c}{2}\] or \[\frac{a}{1}=\frac{b}{-1}=\frac{c}{-2}\] \ Required vector is \[\mathbf{i}-\mathbf{j}-2\mathbf{k}.\]
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