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Let the lines $$\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$$ and $$\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$$ be coplanar and $$\mathrm{P}$$ be the plane containing these two lines. Then which of the following points does NOT lie on P?

JEE · Math · previous-year question

  1. A.$$(0,-2,-2)$$
  2. B.$$(-5,0,-1)$$
  3. C.$$(3,-1,0)$$
  4. D.$$(0,4,5)$$correct

Answer

D. $$(0,4,5)$$

Explanation

$${L_1}:{{x - 1} \over \lambda } = {{y - 2} \over 1} = {{z - 3} \over 2}$$, through a point $${\overrightarrow a _1} \equiv (1,2,3)$$ parallel to $${\overrightarrow b _1} \equiv (\lambda ,1,2)$$ $${L_2}:{{x + 26} \over { - 2}} = {{y + 18} \over 3} = {{z + 28} \over \lambda }$$ through a point $${\overrightarrow a _2} = ( - 26, - 18, - 28)$$ parallel to $${\overrightarrow b _2} = ( - 2,3,1)$$ If lines are coplanar then, $$({\overrightarrow a _2} - {\overrightarrow a _1})\,.\,{\overrightarrow b _1} \times {\overrightarrow b _2} = 0$$ $$ \Rightarrow \left| {\begin{matrix} {27} & {20} & {31} \\ \lambda & 1 & 2 \\ { - 2} & 3 & \lambda \\ \end{matrix} } \right| = 0 \Rightarrow \lambda = 3$$ Vector normal to the required plane $$\overrightarrow n = {\overrightarrow b _1} \times {\overrightarrow b _2}$$ $$ \Rightarrow \overrightarrow n = \left| {\begin{matrix} {\widehat i} & {\widehat j} & {\widehat k} \\ 3 & 1 & 2 \\ { - 2} & 3 & 3 \\ \end{matrix} } \right| = - 3\widehat i - 13\widehat j + 11\widehat k$$ Equation of plane $$ \equiv ((x - 1),(y - 2),(z - 3))\,.\,( - 3, - 13,11) = 0$$ $$ \Rightarrow 3x + 13y - 11z + 4 = 0$$ Checking the option gives (0, 4, 5) does not lie on the plane.

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