Three-dimensional geometry, often referred to as 3D geometry, is a fundamental topic in mathematics that deals with shapes and objects in a three-dimensional space defined by three coordinates: x, y, and z. Unlike two-dimensional geometry, which only considers length and width, three-dimensional geometry incorporates height, adding depth to the study of geometric figures.
Practice questions on three-dimensional geometry typically cover a range of topics such as calculating the direction cosines of a line, finding vector and Cartesian equations of lines and planes, and solving problems involving distances between points and planes.
Important Formula in Three Dimensional Geometry
Distance Formula: The distance between two points A (x1, y1, z1) and B (x2, y2, z2) in a three-dimensional space is presented by:
AB = √[(x2−x1)2+(y2−y1)2+(z2−z1)2]
Section Formula: If A (x1, y1, z1) and B (x2, y2, z2) are two points in a space and let C be a point on the line segment joining A and B such that
- It divides AB internally in the proportion m:n. Then, the coordinates of C are:
C{x, y, z} = {(mx2 + nx1)/(m+n), (my2+ny1)/(m+n), (mz2 + nz1)/(m+n)}.
- If it divides AB externally in proportion m:n. Then, the coordinates of C are:
C{x, y, z} = {(mx2 - nx1)/(m - n), (my2 - ny1)/(m - n), (mz2 - nz1)/(m - n)}
- The equation of a line passing through the two points is r=a+λ(b–a)
The angle θ between two lines whose direction cosines are l1, m1, n1 and l2, m2, n2 is given by:
cos θ = l1 ⋅ l2 + m1 ⋅ m2 + n1 ⋅ n2
- The general equation ax + by + cz + d = 0 represents a plane where a, b and c are constants followed by the condition a, b, c ≠ 0.
Distance of a Plane from a Point
The perpendicular distance of a plane ax + by + cz + d = 0 from a point P (x1, y1, z1) is given by:
Distance = (ax1 + by1 + cz1 + d)/ (a2+b2+c2)
Direction Cosines: If a line forms an angle α, β, γ in the positive direction concerning X-axis, Y-axis and Z-axis, respectively, then cos α, cos β, and cos γ are called its direction cosines.
- If l, m, n are the direction cosines of a line, then l2 + m2 + n2 = 1.
Direction Ratios: Three numbers, say a, b, c, proportional to the direction cosines, say l, m, n of a line, are acknowledged as the direction ratios of the line. Thus a, b, and c are termed the direction ratios of a line provided.
l/a = m/b = n/c
Read More about Direction Cosines and Direction Ratios.
Problem Practice on Three Dimensional Geometry
Problem 1: Find the direction ratios and direction cosines of a point (4, 5, −2).
Solution:
The given point is (4, 5, −2) is represented as a vector a = 4i + 5j − 2k
We know that,
|a| = √[(4)2+(5)2+(−2)2] = √(16 + 25 + 4) = √45 = 3√5
Direction ratios: (a, b, c) = (4, 5, −2)
Direction cosines: (a/√(a2+b2+c2),b/√(a2+b2+c2), c/√(a2+b2+c2) = (4/3√5, 5/3√5, −2/3√5)
Problem 2: What is the equation of a line in three-dimensional geometry, passing through the points (1,3,−2), and (−1,4,3)?
Solution:
The given points are (1,3,−2), and (−1,4,3)
The equation of a line passing through the two points is r=a+λ(b–a)
r = (1i + 3j − 2k) + λ[(−1i + 4j + 3k) − (1i + 3j − 2k)]
⇒ r = (1i + 3j − 2k) + λ[−2i + 1j + 5k]
⇒ xi + yj + zk = (1 − 2λ)i + (3 + λ)j + (−2 + 5λ)k
⇒ (x − 1)i + (y − 3)j + (z + 2)k = −2λi + λj + 5λk
⇒ (x − 1)/−2 = (y − 3)/1 = (z + 2)/5.
Problem 3: If a line makes angles 90°, 135°, 45° with the x, y and z axes respectively, find its direction cosines.
Solution:
Let the direction cosines of the line be l, m, and n.
l = cos 90° = 0
m = cos 135° = -1/√2
n = cos 45° = 1/√2
Hence, the direction cosines of the line are 0, -1/√2, and 1/√2.
Problem 4: Find the angle between the pair of lines given below.
- (x + 3)/3 = (y -1)/5 = (z + 3)/4
- (x + 1)/1 = (y – 4)/1 = (z – 5)/2
Solution:
Given,
(x + 3)/3 = (y -1)/5 = (z + 3)/4
(x + 1)/1 = (y – 4)/1 = (z – 5)/2
The direction ratios of the first line are:
a1 = 3, b1 = 5, c1 = 4
The direction ratios of the second line are:
a2 = 1, b2 = 1, c2 = 2
cos θ = [a1 ⋅ a2 + b1 ⋅ b2 + c1 ⋅ c2]/√[(a12 + b12 + c12)(a22 + b22 + c22)]
⇒ cos θ = [3 . 1 + 5 . 1 + 4 . 2]/√[(32+42+52)(12+12+22)]
⇒ cos θ = 16/√(50 × 16)
⇒ cos θ = 16/[5 × 4√2]
⇒ cos θ = 4/5√2
Problem 5: Show that the lines (x – 5)/7 = (y + 2)/-5 = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.
Solution:
Given lines are: (x – 5)/7 = (y + 2)/-5 = z/1 and x/1 = y/2 = z/3
The direction ratios of the given lines are 7, -5, 1 and 1, 2, 3, respectively.
We know that,
Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular to each other if a1a2 + b1b2 + c1c2 = 0
Therefore, 7(1) + (-5) (2) + 1 (3) = 7 – 10 + 3 =0
Hence, the given lines are perpendicular to each other.
Problem 6: Find the distance between the points A(1,2,3) and B (4,5,6).
Solution:
The distance d between two points is given by the formula: Distance = √[(x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2]
Here, x1 = 1, y1 = 2, z1 = 3 and x2 = 4, y2 = 5, z2 = 6.
Distance = √[(4 - 1)2 + (5 - 1)2 + (6 - 3)2]
⇒ Distance = √(32 + 42 + 32)
⇒ Distance = √(9 + 16 + 9)
⇒ Distance = √34.
Problem 7: Find the distance between the points A(1, 2, 3) and B (4, 5, 8).
Solution:
The distance d between two points is given by the formula: Distance = √[(x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2]
Here, x1 = 1, y1 = 2, z1 = 3 and x2 = 4, y2 = 5, z2 = 8.
Distance = √[(4-1)2+(5-1)2+(8-3)2]
⇒ Distance = √(32+42+52 )
⇒ Distance = √(9+16+25)
⇒ Distance = √50
⇒ Distance = 5√2
Problem 8: How to calculate distant of point A(x, y, z) from Y-axis?
Solution:
Let’s draw a perpendicular line from point A to the Y Axis. The intersection point’s coordinates are B(0, y0, 0). Hence, the required distance AB is
AB = √[(0-x)2 + (y - y0)2 + (0 - z)2]
⇒ AB = x2 + (y - y0)2 + z2
Problem 9: Find the direction cosines if a line makes angles 90°, 45°, and 30°with three axes (x-axis, y-axis, and z-axis).
Solution:
Let’s take the required direction cosines as l, m, and n. The given angles are 90°, 45°, and 30°.
l = cos 90° = 0
m =cos 45° = 12
n = cos 30° = 32
Hence, the required direction cosines are 0, 12, and 32.
Problem 10 : Check whether the given points P (4, 6, -8), Q (2, -4, 6), and R (6, 16, -22) are collinear.
Solution:
We are given the points P (4, 6, -8), Q (2, -4, 6), and R (6, 16, -22).
Using these, the direction ratios of a line joining P and Q are (2-4, -4-6, 6+8), i.e., (-2, -10, 14). The direction ratios of a line joining Q and R are (6-2, 16+4, -22-6), i.e., (4, 20, -28).
Ratios of DRs (Direction ratios) of PQ and QR are -12, -12, -12. Since the DRs seem proportional. So, PQ must be parallel to QR. Therefore, points P, Q, and R are collinear.
Worksheet : Three Dimensional Geometry
Problem 1: Find the angle between the pair of lines given by
- r =3i + 2j - 4k + λ(i + 2j + 2k)
- r = 5i + 2j + μ(3i + 2j + 6k).
Problem 2: Find the direction cosines of a line whose direction ratios are 2, -6, 3.
Problem 3: Find the direction cosines of a line that makes equal angles with the coordinate axes.
Problem 4: Find the angles of triangle ABC whose vertices are A(-1, 3, 2), B (2, 3, 5) and C(3, 5, -2).
Problem 5: Find the angles between the lines whose direction ratios are 3, 2, -6 and 1, 2, 2.
Problem 6: A line makes an angle 60 degree and 45 degrees with the positive direction of x-axis and y-axis respectively. What acute angle does it make with the z-axis?
Problem 7: Show that the lines (x-1)/2 = (y-2)/2 = (z-3)/2 and (x-4)/5 = (y-1)/2 = z intersect each other. Also, find the point of intersection.
Problem 8: Find the equation of the plane which is at a distance 3√3 units from origin and the normal to which is equally inclined to coordinate axis.
Problem 9: Find the angle between the lines whose direction cosines are given by the equation: l+m+n = 0, l2 + m2 + n2 = 0.