A) \[\sqrt{2}\,(3\mathbf{i}+4\mathbf{j})\] or \[-\sqrt{2}\,(3\mathbf{i}+4\mathbf{j})\]
B) \[\sqrt{2}\,(4\mathbf{i}+3\mathbf{j})\] or \[-\sqrt{2}\,(4\mathbf{i}+3\mathbf{j})\]
C) \[\sqrt{3}\,(4\mathbf{i}+5\mathbf{j})\] or \[-\sqrt{3}\,(4\mathbf{i}+5\mathbf{j})\]
D) \[\sqrt{3}\,(5\mathbf{i}+4\mathbf{j})\] or \[-\sqrt{3}\,(5\mathbf{i}+4\mathbf{j})\]
Correct Answer: A
Solution :
Let \[\mathbf{a}=x\mathbf{i}+y\mathbf{j},\] then \[\mathbf{a}\,.\,\mathbf{b}=0\] \[\Rightarrow 4x-3y=0\Rightarrow \frac{x}{3}=\frac{y}{4}\Rightarrow x=3\lambda ,\,\,y=4\lambda ,\,\,\lambda \in R.\] Now \[|\mathbf{a}|\,=\,|\mathbf{b}|\,\Rightarrow {{x}^{2}}+{{y}^{2}}=16+9+25\] \[=9{{\lambda }^{2}}+16{{\lambda }^{2}}=50\] \[\Rightarrow \lambda =\pm \sqrt{2}\]\[\Rightarrow x=\pm 3\sqrt{2},\,\,y=\pm 4\sqrt{2}\] Hence, \[\mathbf{a}=\pm \sqrt{2}(3\mathbf{i}+4\mathbf{j})\].
You need to login to perform this action.
You will be redirected in
3 sec
