Three Dimensional Geometry Question 12
Question 12 - 25 January - Shift 2
If the shortest distance between the line joining the points $(1,2,3)$ and $(2,3,4)$, and the line $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0}$ is $\alpha$, then $28 \alpha^{2}$ is equal to
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Answer: 18
Solution:
Formula: Equation of a straight line , Skew Lines
$ \begin{aligned} & \overrightarrow{{}r}=(i+2 j+3 k)+\lambda(i+j+k) \quad \overrightarrow{{}r}=\overrightarrow{{}a}+\lambda \overrightarrow{{}p} \\ & \overrightarrow{{}r}=(+\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}-\hat{j}) \quad \overrightarrow{{}r}=\overrightarrow{{}b}+\mu \overrightarrow{{}q} \\ & \overrightarrow{{}p} \times \overrightarrow{{}q}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} =\hat{i}+2 \hat{j}-3 \hat{k} \\ & d=|\frac{(\vec{b}-\vec{a}) \cdot(\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|}| \\ & d=|\frac{(-3 \hat{j}-\hat{k}) \cdot(\hat{i}+2 \hat{j}-3 \hat{k})}{\sqrt{14}}| \\ & =|\frac{-6+3}{\sqrt{14}}|=\frac{3}{\sqrt{14}} \\ & \alpha=\frac{3}{\sqrt{14}} \end{aligned} $
Now, $28 \alpha^{2}= 28 \times \frac{9}{14}=18$
