Two intersecting lines lying in plane P1 have equations (x - 1/1)=(y - 3/2)=(z - 4/3) and (x - 1/2)=(y - 3/3)=(z - 4/1). If the equation of plane P2 is 7x-5y+z-6=0 , then the distance between planes P1 and P2 is

Question

Two intersecting lines lying in plane P1 have equations (x - 1/1)=(y - 3/2)=(z - 4/3) and (x - 1/2)=(y - 3/3)=(z - 4/1). If the equation of plane P2 is 7x-5y+z-6=0 , then the distance between planes P1 and P2 is (A) (11/5 √3) (B) (2/√3) (C) (1/√3) (D) (7/5 √3)

Tardigrade Admin · Accepted Answer

Equation of plane P1 is | x-1 y-3 z-4 1 2 3 2 3 1 |=0 (x - 1)(- 7)-(y - 3)(- 5)+(z - 4)(- 1)=0 7x-5y+z+4=0 So, distance between planes P1 &P2 is =|(10/√49 + 25 + 1)|=(10/5 √3) =(2/√3)

Q. Two intersecting lines lying in plane $P_{1}$ have equations $\frac{x - 1}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ and $\frac{x - 1}{2} = \frac{y - 3}{3} = \frac{z - 4}{1} .$ If the equation of plane $P_{2}$ is $7 x - 5 y + z - 6 = 0$ , then the distance between planes $P_{1}$ and $P_{2}$ is

$\frac{11}{5 3}$

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{7}{5 3}$

Equation of plane $P_{1}$ is $∣ ∣ x - 1 12 y - 3 23 z - 4 31 ∣ ∣ = 0$
$(x - 1) (- 7) - (y - 3) (- 5) + (z - 4) (- 1) = 0$
$7 x - 5 y + z + 4 = 0$
So, distance between planes $P_{1} & P_{2}$ is $= ∣ ∣ \frac{10}{49 + 25 + 1} ∣ ∣ = \frac{10}{5 3}$
$= \frac{2}{3}$

Q. Two intersecting lines lying in plane P1​ have equations 1x−1​=2y−3​=3z−4​ and 2x−1​=3y−3​=1z−4​. If the equation of plane P2​ is 7x−5y+z−6=0 , then the distance between planes P1​ and P2​ is

53​11​

3​2​

3​1​

53​7​