The parametric equation of the line passing throught the points $A(3,4,-7)$ and $B(1,-1,6)$ are

The parametric equation of the line passing throught the points $A(3,4,-7)$ and $B(1,-1,6)$ are
  1. $x=1+3 \lambda, \quad y=-1+4 \lambda, \quad z=6-7 \lambda$
  2. $x=-2+3 \lambda, \mathrm{y}=-5+4 \lambda, \quad \mathrm{z}=13-7 \lambda$
  3. $x=3-2 \lambda, y=4-5 \lambda, \quad \mathrm{z}=-7+13 \lambda$
  4. $x=3+\lambda, \quad \mathrm{y}=-1+4 \lambda, \quad \mathrm{z}=-7+6 \lambda$

Solution

Cartesian Equation is $\begin{aligned} & \frac{x-x_{1}}{x_{1}-x_{2}}=\frac{y-y_{1}}{y_{1}-y_{2}}=\frac{z-z_{1}}{z_{1}-z_{2}} \text { i.e. } \\ & \frac{x-3}{3-1}=\frac{y-4}{4+1}=\frac{z+7}{-7-6} \Rightarrow \frac{x-3}{2}=\frac{y-4}{5}=\frac{z+7}{-13}=\lambda \quad \text {...say } \\ \therefore & \frac{x-3}{-2}=\frac{y-4}{-5}=\frac{z+7}{13}=\lambda \Rightarrow x=3-2 \lambda, y=4-5 \lambda, z=-7+13 \lambda \end{aligned}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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