Parametric eqs,Eq. for the plane,Pts of intersection of the line,Pts lie in the same plane,planes para. or per

Question

Lines & Planes in 3-space problems?
Plz guide me in the following questions, step by step & plz write all important formulas, since my Linear Algebra is not good:

10a-Find parametric equation for the line passing through the given points
(5,-2.4), (7, 2,-4). Answer: x=5+t, y=-2+2t, z=4-4t

11a-Find Parametric equations for the line of intersection of the given points
7x-2y+3z=-2 and -3x+y+2z+5=0 Ans.: x=-12-7t, y=-41-23t, z=t

14a-Determine whether the planes are perpendicular
(-2,1,4).(x-1,y,z+3)=0;(1,-2,1).(x+3,y-5,z)=0

16-Line
x=0, y=t, z=t (-Infinity < t < +Infinity)
a-lies in the plane 6x+4y-4z=0
b-is parallel to and below the plane 5x-3y+3z=1
c-is parallel to and above the plane 6x+2y-2y=3

17-Find equation for the plane through (-2,1,7) that is perpendicular to the line x-4=2t, y+2=t, z=-5t. Ans.: 2x+3y-5z+36=0

21-Find an equation for the plane that passes through the points (3,-6,7) and is parallel to the plane 5x-2y+z-5=0
Ans.:5x-2y+z-34=0

22-Find the point of intersection of the line
x-9=-5t, y+1=-t, z-3=t (-Infinity and the plane 2x-3y+4z+7=0
Ans.: (-173/3,-43/3,49/3)

23-Find equation for the plane that contains the line x= -1+3t, y=5+2t, z=2-t and is perpendicular to the plane 2x-4y+2z=9
Ans.: y+2z-9=0

25-Show that the points (-1,-2,-3),(-2,0,1)(-4,-1,-1) and (2,0,1) lie in the same plane.


and if u could provide me ur E-mail address to have future contact in order to ask questions of MATHEMATICS. Thank u

Answer 1

You really need to learn to post only one problem per question.

11a) Find Parametric equations for the line of intersection of the given points
7x-2y+3z=-2 and -3x+y+2z+5=0 Ans.: x=-12-7t, y=-41-23t, z=t

I think you mean the intersection of the given *planes*.

7x - 2y + 3z + 2 = 0
-3x + y + 2z + 5 = 0

The directional vector v, of the line of intersection will be normal to the normal vectors of both planes. Take the cross product of the normal vectors n1 and n2.

v = n1 X n2 = <7, -2, 3> X <-3, 1, 2> = <-7, -23, 1>

Any non-zero multiple of v will also be a directional vector of the line. Multiply by -1.

v = <7, 23, -1>

Now find a point on the line (i.e. a point in both planes).
Let z = 0 and solve for x and y.

7x - 2y + 2 = 0
-3x + y + 5 = 0

Add the first equation to twice the second.

x + 12 = 0
x = -12

Plug back into the second equation and solve for y.

-3(-12) + y + 5 = 0
36 + y + 5 = 0
y = -41

The point of intersection is P(-12, -41, 0).

The equation of the line of intersection can now be written.

L(t) = P + tv
L(t) = <-12, -41, 0> + t<7, 23, -1>
L(t) = <-12 + 7t, -41 + 23t, -t>
where t is a scalar ranging over the real numbers

The parametric form of the equation of the line is:

L(t):
x = -12 + 7t
y = -41 + 23t
z = -t

This answer is mathematically equivalent to your posted answer. The directional vector of the line is just pointing in the opposite direction.
_________

p.s. I will only answer questions that are posted on Yahoo Answers. Any sent directly to me by email will be ignored.

Answer 2

Exactly, you know I was only discussing that, this very night with a can of lager with a broken ring pull. I think the can of Export surmised that there are slightly more cons with exponential functions. AP: I'm not a fan of either but I'll go with bunnies, exponential of course.