Question

Let
\(\begin{array}{l} \vec{a},\vec{b},\vec{c}\ \text{be three non-coplanar vectors such that}\end{array}\)
\(\begin{array}{l} \overrightarrow{a}\times\vec{b}=4\vec{c},\vec{b}\times\vec{c}=9\vec{a}\ \text{and}\ \vec{c}\times\vec{a}=\alpha\vec{b},\alpha>0.\end{array}\)

If
\(\begin{array}{l} \left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|=\frac{1}{36}\end{array}, \) 

then \(α\) is equal to___.

Answer 1

Given, 

\(\begin{array}{l} \overrightarrow{a}\times\overrightarrow{b}=4\cdot\overrightarrow{c}\cdots\left(i\right)\end{array}\)

\(\begin{array}{l} \overrightarrow{b}\times\overrightarrow{c}=9\cdot\overrightarrow{a}\cdots\left(ii\right)\end{array}\)

\(\begin{array}{l} \overrightarrow{c}\times\overrightarrow{a}=\alpha\cdot\overrightarrow{b}\cdots\left(iii\right)\end{array}\)

\(\begin{array}{l}\text{Taking dot products with}\ \overrightarrow{c},\overrightarrow{a},\overrightarrow{b},\ \text{we get}\end{array}\)
\(\begin{array}{l} \overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{c}=\overrightarrow{c}\cdot\overrightarrow{a}=0\end{array}\)
Hence, 
\(\begin{array}{l} \left(i\right)\Rightarrow \left|\overrightarrow{a}\right|\cdot\left|\overrightarrow{b}\right|=4\cdot\left|\overrightarrow{c}\right|\cdots\left(iv\right)\end{array}\)

\(\begin{array}{l} \left(ii\right)\Rightarrow \left|\overrightarrow{b}\right|\cdot\left|\overrightarrow{c}\right|=9\cdot\left|\overrightarrow{a}\right|\cdots\left(v\right)\end{array}\)

\(\begin{array}{l} \left(iii\right)\Rightarrow \left|\overrightarrow{c}\right|\cdot\left|\overrightarrow{a}\right|=\alpha\cdot\left|\overrightarrow{b}\right|\cdots\left(vi\right)\end{array}\)
Multiplying (iv), (v) and (vi)
\(\begin{array}{l} \Rightarrow \left|\overrightarrow{a}\right|\cdot\left|\overrightarrow{b}\right|\cdot\left|\overrightarrow{c}\right|=36\alpha\cdots\left(vii\right)\end{array}\)
Dividing (vii) by (iv)
\(\begin{array}{l} \Rightarrow\ \left|\overrightarrow{c}\right|^2=9\alpha\Rightarrow \left|\overrightarrow{c}\right|=3\sqrt{\alpha}\cdots\left(viii\right)\end{array}\)
Dividing (vii) by (v)
\(\begin{array}{l} \Rightarrow\ \left|\vec{a}\right|^2=4\alpha\Rightarrow\left|\overrightarrow{a}\right|=2\sqrt{\alpha}\end{array}\)
Dividing (viii) by (vi)
\(\begin{array}{l} \Rightarrow\ \left|\overrightarrow{b}\right|^2=36\Rightarrow \left|\overrightarrow{b}\right|=6\end{array}\)
Now, \(\begin{array}{l} 3\sqrt{\alpha}+2\sqrt{\alpha}+6=\frac{1}{36}\Rightarrow \sqrt{\alpha}=\frac{-43}{36}\end{array}\)