If for some R, the lines L_1:x+12=y-2-1=z-11 and L_2:x+2=y+15-=z+11 are coplanar, then the…

Question

If for some $\alpha\in\mathbb R$, the lines $L_1:\dfrac{x+1}{2}=\dfrac{y-2}{-1}=\dfrac{z-1}{1}$ and $L_2:\dfrac{x+2}{\alpha}=\dfrac{y+1}{5-\alpha}=\dfrac{z+1}{1}$ are coplanar, then the line $L_2$ passes through the point:

Accepted Answer

L1 : $${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$$ and

L2 : $${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$$ are coplanar.

$$ \therefore $$ $$\left| {\begin{pmatrix} 1 & 3 & 2  \
   2 & { - 1} & 1  \
   \alpha  & {5 - \alpha } & 1  \

\end{pmatrix} } \right|$$ = 0

$$ \Rightarrow $$ --1(--1 + $$\alpha $$ - 5) + 3(2 - $$\alpha $$) - 2(10 - 2$$\alpha $$ + $$\alpha $$) = 0

$$ \Rightarrow $$ 6 - $$\alpha $$ + 6 - 3$$\alpha $$ + 2$$\alpha $$ - 20 = 0

$$ \Rightarrow $$ --8 --2$$\alpha $$ = 0

$$ \Rightarrow $$ $$\alpha $$ = -4

$$ \therefore $$ Equation of L2 : $${{x + 2} \over { - 4}} = {{y + 1} \over 9} = {{z + 1} \over 1}$$

Check options (2, --10, --2) lies on L2.

Solutions · 1