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p1=(1,-2,1)
p2=(0,-2,3)

Then find the intersection point between this line and the plane which pass by the original point and p1=(2,0,1) and p2=(0,4,0)

can any body help me please, i need to understand how to solve problems like this one.

2007-04-06 10:53:51 · 3 answers · asked by LOST MAN 1 in Science & Mathematics Mathematics

the original point is (0,0,0) as i think

2007-04-06 12:03:57 · update #1

i need more explanation about how to find intersection between two planes or a plane and a line with example please ..........

2007-04-06 12:08:07 · update #2

3 answers

Find the parametric equation for the line which passes thru the two points p1 and p2.

p1 = (1,-2,1)
p2 = (0,-2,3)

Let w be the directional vector of the line.

w = p1p2 = p2 - p1 = <0-1, -2-(-2), 3-1> = <-1, 0, 2>

The equation of the line L can be determined by the directional vector and one point. Let's use p2.

L = p2 + tw = <0,-2,3> + t<-1, 0, 2> = <0 - t, -2 + 0t, 3 + 2t>
L = <-t, -2, 3 + 2t>
where t is a scalar that ranges over the real numbers

Now to get the parametric equations of the line, just break the vector equation of the line into the x, y, and z components.

L:
x = -t
y = -2
z = 3 + 2t
________________________

Then find the intersection point between the line above and the plane which passes thru the original point and
p1 = (2,0,1) and p2 = (0,4,0). I take from the subsequent comments that the original point is the origin (0,0,0).

We have a plane defined by three points. I am relabeling them so that the p1 and p2 of the plane are not confused with the p1 and p2 of the line.

O(0, 0, 0)
Q(2, 0, 1)
R(0, 4, 0)

First, form two vectors u and v, from the three points.

u = OQ = Q - O = <2-0, 0-0, 1-0> = <2, 0, 1>
v = OR = R - O = <0-0, 4-0, 0-0> = <0, 4, 0>

The cross product n of u and v will be normal to both of them and to the plane in which they lie.

n = u X v = <2, 0, 1> X <0, 4, 0> = -4i + 0j + 8k
Any non-zero multiple of n will do as well. Divide by -4.
n = i - 2k = <1, 0, -2>

Now write the equation of plane P with the normal vector n of the plane and one of the points in the plane. Choose point O.

1(x - 0) + 0(y - 0) - 2(z - 0) = 0
x - 2z = 0
____________________________

Find the intersection between the line L and the plane P.

L:
x = -t
y = -2
z = 3 + 2t

P: x - 2z = 0

Plug in the parametric values of the line into the equation of the plane to solve for t.

x - 2y = 0
-t - 2(3 + 2t) = 0
-t - 6 - 4t = 0
-5t = 6
t = -6/5

Now plug that value into the parametric equations of the line to determine the point of intersection between the line and the plane.

L:
x = -t = -(-6/5) = 6/5
y = -2
z = 3 + 2t = 3 + 2(-6/5) = 3/5

The point of intersection is (6/5, -2, 3/5).

2007-04-06 15:53:59 · answer #1 · answered by Northstar 7 · 0 0

The basic concept for this line is to pretend you're starting at p1 and moving in the direction of p2 in one time unit. Let t represent that time unit.

The x distance between the two points is (x2-x1). So, strating at x1, you move (x2-x1) graph units per unit time:

x = x1 + t*(x2-x1)

The y and z directions work the same way:
y = y1 + t*(y2-y1)
z = z1 + t*(z2-z1)

More generally,
p(t) = p1 + t*(p2 - p1).

As for the plane ... do I understand that the plane in question is determined by (1, -2, 1), (2, 0, 1), and (0, 4, 0)? Regardless of the second and third points, the obvious intersection is that very first point, which both the line and plane have in common!

2007-04-06 11:11:13 · answer #2 · answered by norcekri 7 · 0 0

x = a million + 2t y = 2 + 2t z = a million + 4t The development is that the constants a million,2,a million could be between the factors' coordinates and the coefficients 2,2,4 could desire to be proportional to the distinction between the factors' coordinates.

2016-12-08 20:13:43 · answer #3 · answered by ? 4 · 0 0

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