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Question: Answered & Verified by Expert
Under what condition are the two lines
$\mathrm{y}=\frac{\mathrm{m}}{\ell} \mathrm{x}+\alpha, \mathrm{z}=\frac{\mathrm{n}}{\ell} \mathrm{x}+\beta ;$ and $\mathrm{y}=\frac{\mathrm{m}^{\prime}}{\ell^{\prime}} \mathrm{x}+\alpha^{\prime}, \mathrm{z}=\frac{\mathrm{n}^{\prime}}{\ell^{\prime}} \mathrm{x}+\beta^{\prime}$
orthogonal ?
MathematicsThree Dimensional GeometryNDANDA 2008 (Phase 1)
Options:
  • A $\alpha \alpha^{\prime}+\beta \beta^{\prime}+1=0$
  • B $\left(\alpha+\alpha^{\prime}\right)+\left(\beta+\beta^{\prime}\right)=0$
  • C $\ell \ell^{\prime}+\mathrm{mm}^{\prime}+\mathrm{nn}^{\prime}=1$
  • D $\quad \ell \ell^{\prime}+\mathrm{mm}^{\prime}+\mathrm{nn}^{\prime}=0$
Solution:
1580 Upvotes Verified Answer
The correct answer is: $\quad \ell \ell^{\prime}+\mathrm{mm}^{\prime}+\mathrm{nn}^{\prime}=0$
Given two lines are: $\mathrm{y}=\frac{\mathrm{mx}}{\ell}+\alpha, \mathrm{z}=\frac{\mathrm{n}}{\ell} \mathrm{x}+\beta$ and
$\mathrm{y}=\frac{\mathrm{m}^{\prime}}{\ell^{\prime}} \mathrm{x}+\alpha^{\prime}, \mathrm{z}=\frac{\mathrm{n}^{\prime}}{\ell^{\prime}} \mathrm{x}+\beta^{\prime}$
These two lines can be represented as
$\frac{\mathrm{y}-\alpha}{\mathrm{m} / \ell}=\frac{\mathrm{x}-0}{1}=\frac{\mathrm{z}-\beta}{\mathrm{n} / \ell}$ and $\frac{\mathrm{y}-\alpha^{\prime}}{\mathrm{m}^{\prime} / \ell^{\prime}}=\frac{\mathrm{x}-0}{1}=\frac{\mathrm{z}-\beta^{\prime}}{\mathrm{n} 1 / \ell^{\prime}}$
They are orthogonal, if $\frac{\mathrm{m}}{\ell} \times \frac{\mathrm{m}^{\prime}}{\ell^{\prime}}+1 \times 1+\frac{\mathrm{n}}{\ell} \times \frac{\mathrm{n}^{\prime}}{\ell^{\prime}}=-1 \Rightarrow \ell \ell^{\prime}+\mathrm{mm}^{\prime}+\mathrm{nn}^{\prime}=0$

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