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  • Let <img alt="\vec{v}=\alpha \hat{\imath}+2 \hat{\jmath}-3 \hat{k}, \vec{w}=2 \alpha \hat{\imath}+\hat{\jmath}-\hat{k}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cvec%7Bv%7D%3D%5Calpha

Let \vec{v}=\alpha \hat{\imath}+2 \hat{\jmath}-3 \hat{k}, \vec{w}=2 \alpha \hat{\imath}+\hat{\jmath}-\hat{k} and \vec{u} be a vector such that |\vec{u}|=\alpha>0. If the minimum value of the scalar triple product [\vec{u} \vec{v} \vec{w}] is -\alpha \sqrt{3401}, and  |\vec{u} \cdot \hat{\imath}|^2=\frac{m}{n} where m and n are coprime natural numbers, then m+nis equal to

Option: 1

3501


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \alpha & 2 & -3 \\ 2 \alpha & 1 & -1 \end{array}\right| \\ & =\hat{\mathrm{i}}-5 \alpha \hat{\mathrm{j}}-3 \alpha \hat{\mathrm{k}} \\ & [\mathrm{u} \mathrm{v} \mathrm{w}]=\overrightarrow{\mathrm{u}} \cdot \overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}}) \\ & =|\overrightarrow{\mathrm{u}}||\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}}| \cos \theta \\ & \text { since }[\mathrm{u} \mathrm{v} \mathrm{w}] \text { is Least } \Rightarrow \cos \theta=-1 \\ & {[\mathrm{u} \mathrm{v} \mathrm{w}]=\left(|\overrightarrow{\mathrm{u}}| \sqrt{1+25 \alpha^2+9 \alpha^2}\right)(-1)} \end{aligned}

\begin{aligned} & \Rightarrow-\alpha \sqrt{1+34 \alpha^2}=-\alpha \sqrt{3401} \\ & \Rightarrow \alpha^2=100 \\ & \Rightarrow \alpha=10 \\ & \overrightarrow{\mathrm{u}} \text { is parallel to } \overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}} \quad\{\because \alpha>0\} \\ & \overrightarrow{\mathrm{u}}=\lambda(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}}) \\ & \overrightarrow{\mathrm{u}}=\lambda(\hat{\mathrm{i}}-50 \hat{\mathrm{j}}-30 \hat{\mathrm{k}}) \end{aligned}

\begin{aligned} & |\overrightarrow{\mathrm{u}}|=10 \\ & |\lambda| \sqrt{3401}=10 \\ & |\lambda|=\frac{10}{\sqrt{3401}} \quad \overrightarrow{\mathrm{u}}= \pm \frac{10}{\sqrt{3401}}(\hat{\mathrm{i}}-50 \hat{\mathrm{j}}-30 \hat{\mathrm{k}}) \\ & |\overrightarrow{\mathrm{u}} . \hat{\mathrm{i}}|^2=\frac{100}{3401}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \mathrm{m}+\mathrm{n}=100+3401=3501 \end{aligned}

Posted by

Kuldeep Maurya

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