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Mathematics Question on Vector Algebra

Let v⃗=αi^+2j^−3k^,w⃗=2αi^+j^−k^\vec{v}=\alpha \hat{i}+2 \hat{j}-3 \hat{k}, \vec{w}=2 \alpha \hat{i}+\hat{j}-\hat{k} and u⃗\vec{u} be a vector such that |\vec{u}|=\alpha&gt0; If the minimum value of the scalar triple product [u⃗  v⃗  w⃗][\vec{u}\,\, \vec{v}\,\, \vec{w}] is −α3401-\alpha \sqrt{3401}, and ∣u⃗⋅i^∣2=mn|\vec{u} \cdot \hat{i}|^2=\frac{m}{n} where mm and nn are coprime natural numbers, then m+nm+n is equal to _____

Answer

The correct answer is 3501

[u⃗v⃗w⃗]=u⃗.(v⃗×w⃗)[\vec{u}\vec{v}\vec{w}]=\vec{u}.(\vec{v}\times\vec{w})

min.(∣u∣∣v⃗×w⃗∣cosθ)=−α3401min.(|u||\vec{v}\times\vec{w}|cos\theta)=-\alpha \sqrt{3401}

⇒cosθ=−1\Rightarrow cos\theta=-1

∣u∣=α(Given)|u|=\alpha(Given)

∣v⃗×w⃗∣=3401|\vec{v}\times\vec{w}|=\sqrt{3401}

v⃗×w⃗=∣i^j^k^ α2−3 2α1−1∣ \vec{v}\times\vec{w}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ \alpha & 2 & -3 \\\ 2\alpha & 1 & -1 \end{vmatrix}

v⃗×w⃗=i^−5αj^−3αk^\vec{v}\times\vec{w}=\hat{i}-5\alpha\hat{j}-3\alpha\hat{k}

∣v⃗×w⃗∣=1+25α2+9α2=3401|\vec{v}\times\vec{w}|=\sqrt{1+25\alpha^{2}+9\alpha^{2}}=\sqrt{3401}

34α2=340034\alpha ^{2}=3400

α2=100\alpha ^{2}=100

α=10(as α>0)\alpha =10(as \: \alpha>0)

so  u⃗=λ(i^−5αj^−3αk^)so\, \: \vec{u}=\lambda (\hat{i}-5\alpha\hat{j}-3\alpha\hat{k})

u⃗=λ2+25α2λ2+9α2λ \vec{u}=\sqrt{\lambda ^{2}+25\alpha ^{2}\lambda ^{2}+9\alpha ^{2 }\lambda}

α2=λ2(1+25α2+9α2) \alpha ^{2}=\lambda ^{2}(1+25\alpha ^{2}+9\alpha ^{2 })

100=λ2(1+34×100) 100=\lambda ^{2}(1+34\times100)

λ2=1003401=mn\lambda ^{2}=\frac{100}{3401}=\frac{m}{n}