if $a, b, c$ are three linearly dependent vectors then
$$a=i+6j+3k$$ $$b=3i+2j+k$$ $$c=(x+1)i +(y-1)j+k$$ and $|c|=\sqrt{6}$ then $(x+y)$ is
Since the vectors are linearly dependent their scalar triple product will be zero we obtain one equation from there and other from the modulus of $c$ .
I don't think my method is a very good one for a this(MCQ) question as it has to be solved under 3-4 mins.
The answer is
1 and 3
Best Answer
Solve the system $$\alpha[1,6,3]+\beta[3,2,1]=[x+1,y-1,1]$$