In this article, we will discuss the methods of solving problems on 3D geometry. 3D geometry problems consist of various topics, such as problems based on direction ratios and direction cosines, lines in 3D space, planes in 3D space, etc. Direction ratios and direction cosines include problems like; finding how distant a certain point is from a particular axis; finding direction cosines when angles are given or when DRs (direction ratios) are given; and checking whether the given points are collinear. The topic of lines has problems such as finding equations of lines in different circumstances, finding the angle and distance between the lines, and finding the point of intersection between them. The concept of planes in 3D consists of problems like calculating the equations of planes for some specific known features; calculating the distance of the point from a plane and how far one plane is from another if they are parallel; and finding the angle between two planes.

Problems based on direction cosines and direction ratios of a line

P1- How distant is 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, y, 0). Hence, the required distance AB is

AB=0-x2+y-y2+(0-z)2

=x2+02+z2

=x2+z2

P2- 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.

P3- 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.

Problems based on lines in space

P1- Identify the equation of a line progressing through the points L (-2, 2, 4) and M (4, 8, 10) in Cartesian form

Solution: We are given the points L (x1,y1,z1) = (-2, 2, 4) and M (x2,y2,z2) = (4, 8, 10).

The Cartesian equation of a line coursing through these points is 

x-x1x2–x1=y-y1y2–y1=z-z1z2–z1

x+24+2=y-28-2=z-410-4

x+26=y-26=z-46

x+21=y-21=z-41

This is the required equation.

P2- Identify the vector equation of the line passing through the point (10, 4, -8) parallel to vector 5i+2j-7k.

Solution: The given points are (10, 4, -8). Its position vector will be p=10i+4j-8k, and the parallel vector will be q=5i+2j-7k.

Hence, the equation of the line in vector form is-

r=p+cq

r=10i+4j-8k+c(5i+2j-7k)

r=10+5ci+4+2cj-(8+7c)k

P3- The given lines are x-65=y-32=z-27 and x3=y3=z2. Find the angle formed between these two lines. 

Solution: We are given two equations of lines, here x-65=y-32=z-27 and x3=y3=z2.

The equations are given in the Cartesian form. Hence, the angle formed between these lines is- 

cos θ=a1.a2+b1.b2+c1.c2a12+b12+c12.a22+b22+c22

cos θ=5.3+2.3+7.252+22+72.32+32+22

cos θ= 15+6+1425+4+49.9+9+4

cos θ=3578.22

cos θ=351,716

θ=cos-1(351,716)

Problems based on Plane in 3D geometry

P1- If the plane is at a distance of 6 units from the origin and the normal vector coursing through the origin is 4i-5j+6k, then find the plane’s  equation in vector form.

Solution: The normal vector is given as n=4i-5j+6k. Now, the normal unit vector will be-

n=nn

n=4i-5j+6k42+(-5)2+62

n=4i-5j+6k77

It is given that the plane is 6 units distant from the origin.

Hence, the required equation of the plane in vector form is-

r.n=d

r.4i-5j+6k77=6

r.4i-5j+6k=677

P2- A plane progresses through the point with position vector 3i-2j+2k and is perpendicular to the vector 5i–j+3k . Identify the equation of the plane in vector form.

Solution: We are given that the position vector is a=3i-2j+2k, and the normal vector is n=5i–j+3k.

Hence, the plane’s equation in normal form-

r–a.n=0

r.n=a.n

r.5i–j+3k=3i-2j+2k.5i–j+3k

r.5i–j+3k=15+2+6

r.5i–j+3k=23

P3- Determine the plane’s equation whose intercepts are 5, 6, and 7 on the X, Y, and Z-axes.

Solution: We are given that the X-intercept of the plane is a=5, the Y-intercept is b=6, and the Z-intercept is c=7.

Hence, the required equation of the plane is-

xa+yb+zc=1

x5+y6+z7=1

42x+35y+30z210=1

42x+35y+30z=210

P4- The two planes are given as x-2y+z=0 and 2x+y-3z=0. Find the angle formed between these two planes.

Solution: Given equations of planes are  x-2y+z=0 and 2x+y-3z=0.

The angle formed between these two planes is-

cos =a1.a2+b1.b2+c1.c2a12+b12+c12.a22+b22+c22

cos =1.2+-2.1+1.(-3)12+(-2)2+12.22+12+(-3)2

cos =2-2-36.14

cos =384

384

Conclusion

3D geometry deals with the location of points in space,  and these points form lines, planes and many other three dimensional figures. It has many subtopics such as direction ratios and direction cosines, the concept of line in space,  the concept of  the plane, equations of line and plane, distance between two points in space, intersection points of line or plane etc. All these topics have their specific types of problems. These problems serve their uses in daily lives in various ways, for example, maintaining visual graphics, programming for video games, making the design of stairs, construction blueprints etc.