MHTCET 20 Apr 2026 Shift 2 Paper

Concept:Two vectors are mutually perpendicular when their dot product equals zero.Explanation:The given vectors are:$$\vec{A}=4 \hat{i}+n \hat{j}+2 \hat{k}$$and$$\vec{B}=2 \hat{i}+2 \hat{j}-\hat{k}$$For perpendicular vectors, the dot product condition is $\vec{A} \cdot \vec{B} = 0$.Apply the dot product formula by multiplying the corresponding $\hat{i}$, $\hat{j}$, and $\hat{k}$ components:$$\vec{A} \cdot \vec{B} = 4(2) + n(2) + 2(-1)$$Substitute this into the perpendicular condition:$$4(2) + n(2) + 2(-1) = 0$$Simplify each term:$$8 + 2n - 2 = 0$$Combine the constant terms:$$6 + 2n = 0$$Isolate $n$ by moving the constant to the right side:$$2n = -6$$Divide both sides by $2$:$$n = -3$$Hence, for the vectors to be mutually perpendicular, $n$ must equal $-3$.Answer:The value of $n$ is $-3$, so the correct option is C.