Lines and Planes

[sysD]

figured it out.. how do i delete a topic?

Edited September 9, 2012 by sysD

[sysD]

EDIT*************

Disregard the first post, please.

 

 

 

 

I have a few problems relating to lines and planes (mostly vector, scalar, and parametric equations). Would someone please check my answers?

 

Questions:

 

 

1.

Write vector equations and parametric equations for:

a)

the line that passes through A(1,-3,1), parallel to vector u=(2,-2,1)

b)

the line through A(3,0,4) and parallel to x-axis

c)

the line through A(-1,2,1) and B(1,2,1)

 

2.

Determine scalar, vector, and parametric equations for the plane that passes through the points A(1,-2,-0), B(1,-2,2), and C(0,3,2).

 

3.

Write a scalar equation for the plane that contains the point (-1,2,0) and is parallel to the plane (x,y,z)=(1,-2,1)+s(3,1,1)+t(4,-2,1).

 

 

Answers:

 

1.

a)

Vector equation:

(x,y,z)=(1,-3,1)+t(2,-2,1)

Parametric equations:

x= 1+2t

y= -2t-3

z= 1+t

 

b)

Vector equation:

(x,y,z)=(3,0,4)+t(1,0,0)

Parametric equations:

x= 3+t

y= 0

z= 4

 

c)

Vector equation:

(x,y,z)=(-1,2,1)+t(2,0,0)

Parametric equations:

x= 2t-1

y= 2

z= 1

 

 

2.

Vector AB = (0,2,2)

Vector AC = (-1,5,2)

Normal Vector = N = AB x AC = (-10,-2,0)

 

0= -10x-2y+0z+d

Sub in values from Given Point A

0 = -10(1) - 2(-2) + 0(0) + d

-d = -10 +4

-d = -6

d = 6

Therefore, the scalar equation is equivalent to:

-10x - 2y + 6 = 0

 

Vector equation:

(x,y,z) = (1,-2,0)+s(0,0,2)+t(-1,5,2)

 

Parametric equations:

x= 1-t

y= -2+5t

z= 2s+2t

 

 

3.

 

Normal Vector = N = (3,1,1) x (4,-2,1) = (3,1,-10)

 

0 = 3x + y - 10z + d

Sub in values from Given Point

-d = 3(-1) + (2) - 10(0)

-d = -3 + 2

-d = -1

d = 1

Therefore, the scalar equation is equivalent to:

3x + y - 10z + 1 = 0

 

 

 

 

 

 

Are these answers correct? Thanks in advance.

Edited September 10, 2012 by sysD