Question

confirm that the linear equation of the plane
P1 containing both lines L1 and L2 is: 26x + 4y 19z + 70 = 0. What
is the distance between the plane P1 and the origin?

Where:
L1--> x=1-4t, y=7t-5,z=-4t
L2--> x=-4+t,y=-1+3t,z=-2+2t
The lines are not skew

This is part of a larger question so let me know if there's information i may have ommited.

Answer 1

Well to confirm it, you could just show that that equation is satisfied by both lines.

Answer 2

Sorry parametric equations and so on confuse me. do i convert them back into the same form as the equation for the plane and then solve it as a simultaneous equation? I just thought i needed a constant for that. Little confused I'll keep looking for a solution.

Answer 3

If you substitute them in as they are, all the t's and constants should cancel.

Answer 4

the lines are for sure intersecting. Thus, there exists one plane containing the given lines.

the find the equation of the plane,
(a) identify the set of direction numbers of the lines. the sets of direction numbers will serve as the vector representation of the lines

(b) calculate the cross product of the two vectors. the cross product is the normal vector representation of the plane.

Answer 5

(c) choose a point \((x_0,y_0,z_0)\) that lie on any of the given lines. this can be done by
choosing a particular value of \(t\).

(d) if \(<a, b, c>\) is the cross product, then the equation of the plane is
\[\large \qquad a(x−x_0)+b(y−y_0)+c(z−z0)=0\]

Answer 6

Sorry my Internet kicked out for a while. I did what bfsgd recommended and had all of the equations cancel out by substituting in the parametric equations. Does this confirm that the linear equation of the plane contains both L1 and L2?

Now i need to look for how far away P1 is from the origin. Is it just 70 away from the origin?

And i'm unsure of how to identify the set of direction numbers of the lines. Could you elaborate a little, or if the other solution is sufficient i'll just use that.

Answer 7

Yep. If both lines satisfy the equation then the plane must contain both lines.

Answer 8

awesome, is my guess right about p1 being 70 units below the origin, or am i way off?

Answer 9

Is it the shortest distance from the plane to the origin? Since the whole plane cannot be a certain distance from the origin.

Answer 10

"what is the distance between the plane P1 and the origin."

Maybe it's something for my tutor to explain.

Answer 11

the distance asked is the shortest distance of the point to the plane.

Answer 12

N is the cross product,
(x0, y0, z0) is the point on the plane (determined by a particular value of t)

calculate vector OP

calculate the scalar projection of OP onto N (using dot product). the scalar projection is the distance of the origin to the plane P1