MHTCET 23 Apr 2025 Shift 2 Paper

Concept:Use the vector triple product identity $(\bar{u}\times\bar{v})\times\bar{w}=(\bar{u}\cdot\bar{w})\bar{v}-(\bar{v}\cdot\bar{w})\bar{u}$ and simplify using the given magnitudes and dot product.Explanation:First, note that $|\bar{a}|^2=\frac{1}{10}(3^2+0^2+1^2)=\frac{10}{10}=1$.Also, $|\bar{b}|^2=\frac{1}{49}(2^2+3^2+(-6)^2)=\frac{49}{49}=1$.Thus, both $\bar{a}$ and $\bar{b}$ are unit vectors.Their dot product is $\bar{a}\cdot\bar{b}=\frac{3(2)+0(3)+1(-6)}{7\sqrt{10}}=0$.So $\bar{a}$ and $\bar{b}$ are perpendicular.Now take $\bar{u}=\bar{a}$, $\bar{v}=\bar{b}$, and $\bar{w}=\bar{a}+2\bar{b}$.Then $(\bar{a}\times\bar{b})\times(\bar{a}+2\bar{b})=(\bar{a}\cdot(\bar{a}+2\bar{b}))\bar{b}-(\bar{b}\cdot(\bar{a}+2\bar{b}))\bar{a}$.We have $\bar{a}\cdot(\bar{a}+2\bar{b})=|\bar{a}|^2+2\bar{a}\cdot\bar{b}=1+0=1$.And $\bar{b}\cdot(\bar{a}+2\bar{b})=\bar{a}\cdot\bar{b}+2|\bar{b}|^2=0+2=2$.Therefore, the expression becomes $\bar{b}-2\bar{a}$.So the required value is $(2\bar{a}-\bar{b})\cdot(\bar{b}-2\bar{a})=(2\bar{a}-\bar{b})\cdot(-(2\bar{a}-\bar{b}))=-|2\bar{a}-\bar{b}|^2$.Now $|2\bar{a}-\bar{b}|^2=4|\bar{a}|^2+|\bar{b}|^2-4\bar{a}\cdot\bar{b}=4+1-0=5$.Hence the value is $-5$.Answer:Option D. $-5$