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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)
Angle Between Two Lines
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.
FAQs: Three Dimensional Geometry
Define Three Dimensional Geometry.
Three Dimensional Geometry (3D Geometry) is the branch of mathematics that deals with the study of shapes, objects, and figures in three-dimensional space. It involves three coordinates (x, y, and z) to define the position of points, lines, and planes in space.
Why is Three Dimensional Geometry important?
Three Dimensional Geometry is important because it extends the concepts of geometry into the real world, where most objects have three dimensions.
What are the key concepts in Three Dimensional Geometry?
Key concepts in Three Dimensional Geometry include:
- Direction Cosines and Direction Ratios
- Equations of Lines and Planes
- Distance Formula
- Angles Between Lines and Planes
- Skew Lines
How do you find the distance between two points in 3D space?
We can find distance between between two points using distance formula in 3D.
What are Skew Lines?
Skew lines are lines in three-dimensional space that do not intersect and are not parallel.
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