Question

find the ratio in which the line joining (4,8,10)and (6,10,-8) is divided by the plane x+2y+3z-1=0​

Answer 1

Answer:

Equation of line joining two points (a,b,c) and (p,q,r) is given by

$\frac{x-a}{p-a}=\frac{y-b}{q-b}=\frac{z-c}{r-c}$

Equation of  line joining (4,8,10)and (6,10,-8) is given by

$=\frac{x-4}{6-4}=\frac{y-8}{10-8}=\frac{z-10}{-8-10}\\\\=\frac{x-4}{2}=\frac{y-8}{2}=\frac{z-10}{-18}\\\\=\frac{x-4}{1}=\frac{y-8}{1}=\frac{z-10}{-9}=k$

x=k+4, y=k+8, z=-9k+10

The equation of plane is , x+2 y+3 z-1=0​

Substituting the value of , x, y, z in above equation of plane

k+4+2(k+8)+3(-9 k +10)-1=0

k+4+2 k+16-27 k +30-1=0

-24 k+49=0

24 k=49

$k=\frac{49}{24}$

The ratio in which the line joining (4,8,10)and (6,10,-8) is divided by the plane x+2 y+3 z-1=0​ is 49:24.