Advertisements
Advertisements
Question
Find the equation of the plane which passes through the point (3, 4, –1) and is parallel to the plane 2x – 3y + 5z + 7 = 0. Also, find the distance between the two planes
Solution
The required equation parallel to the plane
2x – 3y + 5z + 7 = 0 .........(1)
2x – 3y + 5z + λ = 0 ..........(2)
This passes through (3, 4, -1)
(2) ⇒ 2(3) – 3(4) + 5(–1) + λ = 0
6 – 12 – 5 + 1 = 0
λ = 11
(2) ⇒ The required equation is 2x – 3y + 5z + 11 =0 ........(3)
∴ Now, distance between the above parallel lines (1) and (3)
a = 2
b = – 3
c = 5
d1 = 7
d2 = 11
Distance = `|("d"_1 - "d"_2)/sqrt("a"^2 + "b"^2 + "c"^2)|`
= `|(7 - 11)/sqrt(2^2 + 3^2 + 5^2)|`
= `|4/sqrt(38)|`
d = `4/sqrt(38)`
APPEARS IN
RELATED QUESTIONS
Find the equation of the plane passing through the line of intersection of the planes `vec"r"*(2hat"i" - 7hat"j" + 4hat"k")` = 3 and 3x – 5y + 4z + 11 = 0, and the point (– 2, 1, 3)
Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x – y + z = 3 and at a distance `2/sqrt(3)` from the point (3, 1, –1)
Find the angle between the line `vec"r" = (2hat"i" - hat"j" + hat"k") + "t"(hat"i" + 2hat"j" - 2hat"k")` and the plane `vec"r"*(6hat"i" + 3hat"j" + 2hat"k")` = 8
Find the angle between the planes `vec"r"*(hat"i" + hat"j" - 2hat"k")` = 3 and 2x – 2y + z = 2
Find the length of the perpendicular from the point (1, – 2, 3) to the plane x – y + z = 5
Find the point of intersection of the line with the plane (x – 1) = `y/2` = z + 1 with the plane 2x – y – 2z = 2. Also, the angle between the line and the plane
Find the co-ordinates of the foot of the perpendicular and length of the perpendicular from the point (4, 3, 2) to the plane x + 2y + 3z = 2.
Choose the correct alternative:
If `vec"a" = hat"i" + hat"j" + hat"k", vec"b" = hat"i" + hat"j", vec"c" = hat"i"` and `(vec"a" xx vec"b")vec"c" - lambdavec"a" + muvec"b"` then the value of λ + µ is
