Let the lines [(x-1)/()=(y-2)/(1)=(z-3)/(2)] and [(x+26)/(-2)=(y+18)/(3)=(z+28)/()] be cop…

Question

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?

Accepted Answer

$${L_1}:{{x - 1} \over \lambda } = {{y - 2} \over 1} = {{z - 3} \over 2}$$,

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{pmatrix} {27} & {20} & {31}  \
   \lambda  & 1 & 2  \
   { - 2} & 3 & \lambda   \

\end{pmatrix} } \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{pmatrix} {\widehat i} & {\widehat j} & {\widehat k}  \
   3 & 1 & 2  \
   { - 2} & 3 & 3  \

\end{pmatrix} } \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.

Solutions · 1