sysD Posted September 9, 2012 Share Posted September 9, 2012 (edited) figured it out.. how do i delete a topic? Edited September 9, 2012 by sysD Link to comment Share on other sites More sharing options...

sysD Posted September 10, 2012 Author Share Posted September 10, 2012 (edited) 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 Link to comment Share on other sites More sharing options...

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