Q56 / 120 · VITEEE 2015 · Mathematics

The two lines $x=m y+n, z=p y+q$ and $x=m^{\prime} y+n^{\prime}$, $z=p^{\prime} y+q^{\prime}$ are perpendicular to each other, if

  • A
    $\mathrm{mm}^{\prime}+p p^{\prime}=1$
  • B
    $\frac{m}{m^{\prime}}+\frac{p}{p^{\prime}}=-1$
  • C
    $\frac{m}{m^{\prime}}+\frac{p}{p^{\prime}}=1$
  • Correct answer
    D
    $\mathrm{mm}^{\prime}+p p^{\prime}=-1$

Solution

Given lines are $x=m y+n, z=p y+q$ and $x=m^{\prime} y+n^{\prime}, z=p^{\prime} y+q^{\prime}$ Above equations can be rewritten as $\frac{x-n}{m}=\frac{y-0}{1}=\frac{z-q}{p}$ and $\frac{\mathrm{x}-\mathrm{n}^{\prime}}{\mathrm{m}^{\prime}}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-\mathrm{q}^{\prime}}{\mathrm{p}^{\prime}}$ Lines will be perpendicular, if $\mathrm{mm}^{\prime}+1+\mathrm{pp}^{\prime}=0$ $\Rightarrow \mathrm{mm}^{\prime}+\mathrm{pp}^{\prime}=-1$

Hence, the correct answer is option: 'D'.

Relevant Questions