Practice Questions on Three Dimensional Geometry (original) (raw)

Last Updated : 23 Jul, 2025

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 = √[(x 2 −x 1 ) 2 +(y 2 −y 1 ) 2 +(z 2 −z 1 ) 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

**C{x, y, z} = {(mx 2 **+ nx 1 )/(m+n), (my 2 +ny 1 )/(m+n), (mz 2 **+ nz 1 )/(m+n)}.

**C{x, y, z} = {(mx 2 **- nx 1 )/(m - n), (my 2 **- ny 1 )/(m - n), (mz 2 **- nz 1 )/(m - n)}

**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 = l 1 **⋅ l 2 **+ m 1 **⋅ m 2 **+ n 1 **⋅ n 2

**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 = (ax 1 + by 1 + cz 1 + d)/ (a 2 +b 2 +c 2 )

**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.

**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.

**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 = [a 1 **⋅ a 2 **+ b 1 **⋅ b 2 + c 1 ⋅ c 2 ]/√[(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

**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.