Practice Questions on Three Dimensional Geometry

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.