Problems on Lines in 3D with Detailed Solutions
Problems on lines in 3D with detailed solutions are presented.
Questions

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 P_{0}(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 >  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>
equality of vector components of the above vector equation give: x = 1  t , y = 2  4 t and z = 3  2 t

Solution
Write the symmetric equations form of the line: (x  3)/ 3 = (y  4) / (1) = (z  5) / 2
Point A: Substitute x, y and z in the symmetric equations by the coordinates of point A(3 , 4 , 4).
(x  3)/ 3 = (3  3)/ 3 = 0
(y  4) / (1) = (4  4) / (1) = 0
(z  5) / 2 = (4  5) / 2 =  1 / 2
Last expression is not equal to the first two, hence A is not on the line.
Point B: Substitute x, y and z in the symmetric equations by the coordinates of point B(0 , 5 , 3).
(x  3)/ 3 = (0  3)/ 3 =  1
(y  4) / (1) = (5  4) / (1) = 1
(z  5) / 2 = (3  5) / 2 =  1
All expressions are equal, hence B is on the line.
Point C: Substitute x, y and z in the symmetric equations by the coordinates of point B(6 , 3 , 7).
(x  3)/ 3 = (6  3)/ 3 = 1
(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
Write given line x = 2 t + 5, y = 4 t and z = t + 3. in symmetric form:
(x  5) / 2 = y / (4) = (z  3) / (1)
The direction vector is : <2 ,  4 ,  1>
The line through point P(3 , 5 , 2) is parallel to the given line and hence they have the same direction vector. Hence the vector equation of the line through P:
x = 3 + 2 t , y = 5  4 t , z = 2  t

Solution
Line L (to find) is perpendicular to lines L1 and L2 is therefore perpendicular to their direction vectors d1 and d2 respectively. Hence the cross product of d1 and d2 gives the direction vector of the line L.
From symmetric equation of L1: d1 = <3 , 4 , 4>
Write equation of L2 in symmetric form: (x + 4) / 3 = (y  6) / (1) = z / 5
Direction vector of L2 ; d2 = <3 , 1 , 5>
d direction vector of L may be taken as the cross product of: d = d1 × d2 = <3 , 4 , 4> × <3 , 1 , 5> = <16 , 3 , 9>
Equation for L through point 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
Write the equations of line L2 in parametric form using the parameter s as follows:
x = 4 s + 7 , y = 2 s  2 , z = 3 s + 2
Let A(x , y , z) be the point of intersection of the two lines. To be a point of intersection, the coordinates of A must satisfy the equations of both lines simultaneously. Hence
x coordinates equal gives equation: 2 t  1 = 4 s + 7
y coordinates equal gives equation: 3 t + 2 = 2 s  2
z coordinates equal gives equation: 4 t 3 = 3 s + 2
Rewrite the first two equations in t and s as follows:
2 t  4 s = 8 and  3 t  2 s =  4
Solve for t and s to get: t = 2 and s = 1
To have intersection, t = 2 and s = 1 must also be solution to the third equation 4 t 3 = 3 s + 2.
Check: 4(2)  3 = 5 and  3(1) + 2 = 5 and t = 2 and s =  1 are solutions to all three equations.
The point of intersection is obtained by using one of the two parametric equations of L1 and L2.
Using L1:Set t = 2 in the equations: x = 2t  1 , y = 3 t + 2 and z = 4 t 3 to obtain the coordinates of the point of intersection: x = 3 , y =  4 and z = 5
Using L2 (to check) : x = 4 s + 7 , y = 2 s  2 , z = 3 s + 2 , s =  1 , gives the coordinates of the point of intersection: x = 3 , y =  4 and z = 5

Solution
Let d1 and d2 be the direction vectors of L1 and L2.
For L1: d1 = <2 ,  2 ,  4>
For L2 : d2 = <6 , 2 , 2>
The angle θ between the lines L1 and L 2 is equal to the angle between their direction vectors d1 and d1 which is given by
θ = arccos(d1·d2 / d1 d2)
where d1·d2 is the scalar product of vectors d1 and d2; d1 is the magnitude of vector d1 and d2 is the magnitude of vector d2.
d1·d2 = (2)(6)+(2)(2)+(4)(2) = 0
d1 and d2 are both non zero, hence
θ = arccos(0) = 90°
If the lines intersect, they make an angle of 90°. If they do not intersect, the intersecting lines parallel to these two lines make an angle of 90°.

Solution
One way to approach this problem is to show that the two lines L1 and L2 have two points in common. Rewrite the two equations in parametric form as:
L1: x = 2  t , y = 2 t , z =  1 + 4 t
L2: x = 1  2 t , y = 2 + 4 t , z = 3 + 8 t
Point P1 on L1: (2 , 0 , 1)
Point P2 on L2: (1 , 2 , 3)
Check that P1(2 , 0 , 1) is on L2 using the symmetric equations of L2: (2  1) / (2) = (0  2) / 4 = (1  3) / 8 =  1 / 2 , all terms equal P1 is on L2.
Check that P2(1 , 2 , 3) is on L1 using the symmetric equations of L1: (1  2) / ( 1) = 2 / 2 = (3 + 1) / 4 = 1 , all terms equal, P2 is on L1
P1 is on L1 and L2 and P2 is also on L2 and L1. The two equations are of the same line.

Solution
According to the equation of the line, point P(2,3,0) is on the line. The vector direction is d = <2 ,3 , 1>
The distance D between the line and point P_{0}(1 ,  2 , 3) is given by 1
D =  P_{0}P × d  / d
P_{0}P = <1 , 5 , 3> , d = <2 , 3 , 1>
P_{0}P × d = < 14 , 5 , 13 >
 P_{0}P × d  = √(14^{ 2} + 5^{ 2} + 13^{ 2}) = √390
d = √ ((2)^{ 2} + 3^{ 2} + 1^{ 2}) = √ 14
D = √390 / √ 14 = √195 / √ 7

Solution
Points P1(2 , 0, 1) and P2(1 , 2 , 3) are points on L1 and L2 respectively. d1 = < 1 , 4 , 4> and d2 = <5 , 2 , 2> are the direction vectors of L1 and L2
The shortest distance D between the two lines is given by 1
D =  n · P1 P2  / n
where P1 P2 is the vector defined by points P1 and P2 and n is the cross product of d1 and d2.
n = d1 × d2 = < 1 , 4 , 4> × <5 , 2 , 2> = <0 , 18 , 81>
P1 P2 = < 1 ,  2 , 4 >
D =  <0 , 18 , 18> · < 1 ,  2 , 4 >  / √(0^{2} + 18^{2} + 18^{2} ) = √ 2

Solution
Direction vectors must be proportional < 10 , b , 4 > = k <  5 , 2 ,  2 >
10 =  5 k , k = 2
4 = k (2) , k =  2
b = k(2) = 2 (2) = 4

Solution
a) Equation of line to be found:= < 1 , 2 , 3 > + t < a , b , c >
Perpendicular to := <  3 , 3 , 1 > + m <1 , 1 , 1 >
Apply conditions to find a, b and c.
Intersection gives 3 equations:
1 + a t =  3 + m
2 + b t = 3 + m
3 + c t = 1 + m
add sides of all three equations to obtain
1  2 + 3 +t (a + b + c) = 1 + 3m
Direction vectors < a , b , c > and <1 , 1 , 1 > are orthogonal therefore their dot product is zero.
a + b + c = 0
The above equation 1  2 + 3 +t (a + b + c) = 1 + 3m becomes 3m = 3
Solve for m:
m = 1
Put m = 1 in the first of the three equations: 1 + a t =  3 + m to find a t =  3
Let a = 1 to get t =  3
Use the equation to find b: 2 + b t = 3 + m , 2 + b(3) = 3 + 1 , b =  6 / 3 =  2
Use the equation to find c: 3 + c t = 1 + m , 3 + c(3) =  1 + 1 , c = 1
Equation of line through the point (1 , 2 , 3) : < x , y , z > = < 1 , 2 , 3 > + t <1 , 2 , 1 >
b) Point of intersection
Set t = 3 in the equation of the line found: < x , y , z > = < 1 , 2 , 3 > + t <1 , 2 , 1 > = < 1 , 2 , 3 >  3<1 , 2 , 1 > = < 2 , 4 , 0 >
To check, set m = 1 in the given equation of the line.
< x , y , z > = <  3 , 3 , 1 > + m <1 , 1 , 1 > = <  3 , 3 , 1 > + 1 <1 , 1 , 1 > = < 2 , 4 , 0 >
The is a point of intersection given by: < 2 , 4 , 0 >
Middle School Maths (Grades 6, 7, 8, 9)  Free Questions and Problems With Answers
Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers
Home Page