A) \[\lambda \mathbf{a}\]
B) \[\lambda \mathbf{b}\]
C) \[\lambda \mathbf{c}\]
D) 0
Correct Answer: D
Solution :
Let \[\mathbf{a}+2\mathbf{b}=x\mathbf{c}\] and \[\mathbf{b}+3\mathbf{c}=y\mathbf{a},\] then \[\mathbf{a}+2\mathbf{b}+6\mathbf{c}=(x+6)\mathbf{c}\]and \[\mathbf{a}+2\mathbf{b}+6\mathbf{c}=(1+2y)\mathbf{a}\] So, \[(x+6)\mathbf{c}=(1+2y)\mathbf{a}\] Since \[\mathbf{a}\] and \[\mathbf{c}\] are non-zero and non-collinear, we have \[x+6=0\] and \[1+2y=0\] i.e., \[x=-6\] and \[y=-\frac{1}{2}.\] In either case, we have \[\mathbf{a}+2\mathbf{b}+6\mathbf{c}=\mathbf{0}\].You need to login to perform this action.
You will be redirected in
3 sec