JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 24)
If the shortest distance between the line joining the points (1, 2, 3) and (2, 3, 4), and the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {{z - 2} \over 0}$$ is $$\alpha$$, then 28$$\alpha^2$$ is equal to ____________.
Answer
18
Explanation
Points $(1,2,3)$ and $(2,3,4)$
$L_{1}: \frac{(x-1)}{1}=\frac{(y-2)}{1}=\frac{(2-3)}{1}$
$L_{2}: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0}$
$\vec{b}_{1}=\hat{i}+\hat{j}+\hat{k}$
$\vec{b}_{2}=2 \hat{i}-\hat{j}+0 \hat{k}$
$$ \begin{aligned} & \overrightarrow{a_{1}}-\overrightarrow{a_{2}}=0 \hat{i}-3 \hat{j}-\hat{k} \\\\ & d=\left|\frac{\left(\bar{a}_{1}-\bar{a}_{2}\right) \cdot\left(n_{1} \times n_{2}\right)}{\left|n_{1} \times n_{2}\right|}\right| \\\\ &=\left|\frac{6-3}{\sqrt{9+1+4}}\right|=\frac{3}{\sqrt{14}}=\alpha \\\\ & 28 \alpha^{2}=\frac{28 \times 9}{14}=18 \end{aligned} $$
$L_{1}: \frac{(x-1)}{1}=\frac{(y-2)}{1}=\frac{(2-3)}{1}$
$L_{2}: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0}$
$\vec{b}_{1}=\hat{i}+\hat{j}+\hat{k}$
$\vec{b}_{2}=2 \hat{i}-\hat{j}+0 \hat{k}$
$$ \begin{aligned} & \overrightarrow{a_{1}}-\overrightarrow{a_{2}}=0 \hat{i}-3 \hat{j}-\hat{k} \\\\ & d=\left|\frac{\left(\bar{a}_{1}-\bar{a}_{2}\right) \cdot\left(n_{1} \times n_{2}\right)}{\left|n_{1} \times n_{2}\right|}\right| \\\\ &=\left|\frac{6-3}{\sqrt{9+1+4}}\right|=\frac{3}{\sqrt{14}}=\alpha \\\\ & 28 \alpha^{2}=\frac{28 \times 9}{14}=18 \end{aligned} $$
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