If the lines x-11=y-22=z+31 and x-a2=y+23=z-31 intersect at the point P, then the distance…

Question

If the lines $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z+3}{1}$ and $\dfrac{x-a}{2}=\dfrac{y+2}{3}=\dfrac{z-3}{1}$ intersect at the point $P$, then the distance of the point $P$ from the plane $z=a$ is:

Accepted Answer

$${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1} = \lambda $$ (say)

\& $${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1} = \mu $$ (say)

$$\therefore$$ $$\lambda  + 1 = 2\mu  + a$$ ...... (i)

$$2\lambda  + 2 = 3\mu  - 2$$ ..... (ii)

$$\lambda  - 3 = \mu  + 3$$ .... (iii)

By (i) \& (ii)

$$ \Rightarrow 3\mu  - 2 = 4\mu  + 2a + 2$$

$$\mu=-2(1+a)$$ \& $$\lambda=5-3a$$

Put $$\lambda$$ \& $$\mu$$ in (iii) we get

$a=-9$

$$\mu=16$$

$$\lambda=22$$

$$\therefore$$ Point of intersection $$\equiv(23,46,19)$$

Distance from $z=-9$ is 28

Solutions · 1