A line L1 is drown between two points P1(3,4,7) and P2(5,6,1) and L2 drown between P3(1,5,-2)
and P4(2,9,0) identify the relation between the two lines,
Answers
Answer:
Write the equation of the line given in vector form by < x , y , z > = < -2 , 3 , 0 > + t < 3 , 2 , 5 > into parametric and symmetric forms.
Find the symmetric form of the equation of the line through the point P(1 , - 2 , 3) and parallel to the vector n = < 2, 0 , -3 >.
Find the parametric equations of the line through the two points P(1 , 2 , 3) and Q(0 , - 2 , 1).
Which of the points A(3 , 4 , 4) , B(0 , 5 , 3) and C(6 , 3 , 7) is on the line with the parametric equations x = 3t + 3, y = - t + 4 and z = 2t + 5?
Find the parametric equations of the line through the point P(-3 , 5 , 2) and parallel to the line with equation x = 2 t + 5, y = -4 t and z = -t + 3.
Find the equation of a line through P(1 , - 2 , 3) and perpendicular to two the lines L1 and L2 given by:
L1: (x - 2) / 3 = (y + 1) / (-4) = (z + 9) / 4
L2: x = 3t - 4 , y = - t + 6 and z = 5t .
Find the point of intersection of the lines L1 and L 2 in 3D defined by:
L1 (in parametric form): x = 2t - 1 , y = -3 t + 2 and z = 4 t -3
L2 (in symmetric form) : (x - 7) / 4 = (y + 2) / 2 = (z - 2)/(-3)
Find the angle between the lines L1 and L 2 with symmetric equations:
L1: (x - 1) / 2 = (y + 2) / (-2) = z / -(4)
L2 = (x + 3) / 6 = (y + 2) / 2 = (z - 1) / 2
Show that the symmetric equations given below are those of the same line.
L1: (x - 2) / (- 1) = y / 2 = (z + 1) / 4
L2 = (x - 1) / (-2) = (y - 2) / 4 = (z - 3) / 8
Find distance between point P0(1 , - 2 , 3) and the line with vector equation: < x , y , z > = < 2 , 3, 0 > + t < -2 , 3 , 1 >
Find the shortest distance between the two lines L1 and L2 defined by their equations::
L1: = < 2 , 0 , -1 > + t < -1 , 4 , -4 >
L2: = < 1 , - 2 , 3 > + m < - 5 , 2 , - 2 >
Find value of b so that the lines L1 and L2 given by their equations below are parallel.
L1: = < 2 , 0 , -1 > + t < 10 , b , 4 >
L2: = < 1 , - 2 , 3 > + m < - 5 , 2 , - 2 >
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Find the equation of a line through the point P(1 , -2 , 3) and intersects and is perpendicular to the line with parametric equation x = - 3 + t , y = 3 + t , z = -1 + t. Find the point of intersection of the two lines.
Solutions to the Above Questions
Solution
Given: < x , y , z > = < -2 , 3 , 0 > + t < 3 , 2 , 5 >
equality of vector components of the above vector equation give: x = - 2 + 3 t , y = 3 + 2 t and z = 0 + 5 t
Solve for t each of the above: t = (x + 2) / 3 , t = (y - 3) / 2 and t = z / 5
All equal to t, hence the symmetric form of the equation: (x + 2) / 3 = (y - 3) / 2 = z / 5
Solution
< x , y , z > = <1, -2 , 3> + t <2 , 0 ,-3>
equality of vector components of the above vector equation give: x = 1 + 2 t , y = - 2 and z = 3 - 3 t
Solve for t each of the above: t = (x - 1) / 2 and t = (z - 3) / - 3
Symmetric form of the equation: (x - 1) / 2 = (z - 3) / - 3 and y = -2
Solution
Direction vector PQ = <0 - 1 , - 2 - 2 , 1 - 3 > = <-1 , -4 , -2>
< x , y , z > = <1 , 2 , 3> + t <-1 , -4 , -2>
(y - 4) / (-1) = (3 - 4) / (-1) = 1
(z - 5) / 2 = (7 - 5) / 2 = 1
All expressions are equal, hence C is on the line.
Solution
P(1 , - 2 , 3) and parallel to vector d is given by:
< x , y , z > = <1 , - 2 , 3> + t <-16 , -3 , 9>
Solution
L1: x = 2t - 1 , y = -3 t + 2 and z = 4 t -3
Answer:
all of the above
Explanation:
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