1)find the angle in radians between the planes -2x+z=1 and 3y+z=1.
2)find the distance b/w the point (4,2,1) to the line x=0, y=2+4t, z=1+4t.
3)find the distance from the point (0,5,1) to the plane -3x-3y-1z=-4.
2007-10-18 15:11:16 · 2 answers · asked by vntraderus88 1 in Science & Mathematics ➔ Mathematics
1) Find the angle in radians between the planes
-2x + z = 1 and 3y + z = 1.
The dihedral angle between the planes is equal to the angle between the normal vectors n1 and n2, to the planes.
n1 = <-2, 0, 1>
n2 = <0, 3, 1>
Take the dot product of n1 and n2 and calculate their magnitudes.
n1 • n1 = <-2, 0, 1> • <0, 3, 1> = 0 + 0 + 1 = 1
|| n1 || = √[(-2)² + 0² + 1²] = √(4 + 0 + 1) = √5
|| n2 || = √[0² + 3² + 1²] = √(0 + 9 + 1) = √10
Another definition of the dot product is:
n1 • n1 = || n1 || || n2 || cosθ
where θ is the angle between the vectors
cosθ = (n1 • n1) / (|| n1 || || n2 ||)
cosθ = 1 / [(√5)(√10)] = 1/√50 = 1/(5√2)
θ = arccos[1/(5√2)] ≈ 1.4288993 radians
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2) Find the distance h, between the point P(4,2,1) and the
line x=0, y=2+4t, z=1+4t.
The directional vector v, of the line is:
v = <0, 4, 4>
The magnitude of v is:
|| v || = √(0² + 4² + 4²) = √(0 + 16 + 16) = √32 = 4√2
When t = 0 a point Q on the line is:
Q(0,2,1)
Let the vector u be:
u = QP =
= <4-0, 2-2, 1-1> = <4, 0, 0>
Take the cross product of u and v.
u X v = <4, 0, 0> X <0, 4, 4> = <0, -16, 16>
The magnitude of the cross product of u and v is:
|| u X v || = √[0² + (-16)² + 16²] = √(0 + 256 + 256)
|| u X v || = √512 = 16√2
The magnitude of the cross product of u and v can also be expressed as:
|| u X v || = || u || || v || sinθ = || u || || v || (h / || u ||)
where θ is the angle between the vectors
|| u X v || = || u || || v || (h / || u ||) = || v || h
h = || u X v || / || v || = 16√2 / (4√2) = 4
The distance from the P(4, 2, 1) to the line is 4.
p.s. We could have done this quicker if we had noticed that vectors u and v were perpendicular.
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3) Find the distance from the point P(0,5,1) to the
plane -3x - 3y - 1z = -4.
Rewrite the formula for the plane by multiplying thru by -1.
3x + 3y + z = 4
3x + 3y + z - 4 = 0
Now use the distance formula between a point and a plane.
distance = |3*0 + 3*5 + 1*1 - 4| / √(3² + 3² + 1²)
distance = 12/√19
2007-10-18 18:46:07 · answer #1 · answered by Northstar 7 · 2⤊ 0⤋
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2016-10-07 04:53:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋