MHTCET 11 May 2024 Shift 2 Paper

Concept:Two vectors are collinear when one is a non-zero scalar multiple of the other.For a vector to be equal to both $\lambda \bar{a}$ and $\mu \bar{c}$ with $\bar{a}$ and $\bar{c}$ non-collinear, the only possible value of the vector is $\bar{0}$.Explanation:Since $\bar{a}+2\bar{b}$ is collinear with $\bar{c}$, there exists a non-zero scalar $m$ such that:$$\bar{a}+2\bar{b}=m\bar{c}$$Since $\bar{b}+3\bar{c}$ is collinear with $\bar{a}$, there exists a non-zero scalar $n$ such that:$$\bar{b}+3\bar{c}=n\bar{a}$$Now consider the vector $\bar{a}+2\bar{b}+6\bar{c}$.Using the first relation:$$\bar{a}+2\bar{b}+6\bar{c}=(m+6)\bar{c}$$Using the second relation:$$\bar{a}+2\bar{b}+6\bar{c}=\bar{a}+2(\bar{b}+3\bar{c})=\bar{a}+2n\bar{a}=(1+2n)\bar{a}$$Thus the same vector equals both $(m+6)\bar{c}$ and $(1+2n)\bar{a}$.Since $\bar{a}$ and $\bar{c}$ are not collinear, both scalar coefficients must be zero.$$m+6=0 \quad \text{and} \quad 1+2n=0$$Therefore, the vector becomes $\bar{0}$.Answer:$$\bar{a}+2\bar{b}+6\bar{c}=\bar{0}$$The correct option is D: $\bar{0}$.