The image of the line x-13=y-31=z-4-5 in the plane 2x-y+z+3=0 is the line:

Question

The image of the line $\dfrac{x-1}3=\dfrac{y-3}1=\dfrac{z-4}{-5}$ in the plane $2x-y+z+3=0$ is the line:

Accepted Answer

$${{a - 1} \over 2} = {{b - 3} \over { - 1}} = {{c - 4} \over 1} = \lambda \left( {let} \right)$$

$$ \Rightarrow a = 2\lambda  + 1,\,\,b = 3 - \lambda ,\,\,c = 4 + \lambda $$

$$p = \left( {{{a + 1} \over 2},{{b + 3} \over 2},{{c + 4} \over 2}} \right)$$

$$ = \left( {\lambda  + 1,{{6 - \lambda } \over 2},{{\lambda  + 8} \over 2}} \right)$$

$$\therefore$$ $$2\left( {\lambda  + 1} \right) - {{6 - \lambda } \over 2} + {{\lambda  + 8} \over 2} + 3 = 0$$

$$3\lambda  + 6 = 0 \Rightarrow \lambda  =  - 2$$

$a =  - 3,b = 5,c = 2$

Required line is $${{x + 3} \over 3} = {{y - 5} \over 1} = {{z - 2} \over { - 5}}$$

Solutions · 1