3D geometry and vectors were introduced in Maths to help students grasp different types of shapes and figures. Almost all items in the real world have three dimensions. Many household items, for example, contain 3D geometry, such as pens, laptops, windows, culinary utensils, and so on. Students have studied the fundamentals of three-dimensional geometry in class 11. In class 12, the advanced version of 3-D geometry is taught. Vector algebra will now be applied to three-dimensional geometry.
The goal of this 3-dimensional geometry approach is to create a simple yet elegant study structure. Important concepts covered in three-dimensional geometry for Class 12 include direction cosine and direction ratios of a line connecting two points. The notes are based on the term 2 CBSE and NCERT syllabuses.
Students can review Maths Chapter 11 (Three-dimensional geometry) with notes created according to the most recent exam pattern. The students will also study the equations of lines and planes in space under various conditions, the angle between a line and a plane, and the angle between two lines, among other things. To better understand the material, you should practise the problems based on the NCERT solutions for class 12 Maths. This chapter has great importance in exams like JEE 2022, JEE Advanced 2022, National Engineering Olympiad, and other olympiads in India.
Important Notes
A student preparing for CBSE 12th Maths term two examinations ought to study the notes mentioned below in this article. These notes are created by subject experts and toppers who are well versed with the examination pattern. Since this term’s paper is subjective, there are chances that at least two questions out of four long answer-type questions will come from this chapter. Hence it is important to go through each note mentioned in the article.
Direction Cosine and Direction Ratio
The direction cosine and direction ratios are the first things taught in CBSE 12th 3D geometry. These can be understood by making a line pass through the origin (0,0). Let angles formed by the line and axes be α, β, γ with x, y, and z-axes. Therefore finding the cosine of these angles will give us the direction cosine of the line. For now, let us consider the
x = ka
y = kb
z = kc
xa = yb = zc = k
Therefore the direction cosines will be formulated as:
m = ± a2/ √(a2 + b2 + c2)
m = ± b2/ √(a2 + b2 + c2)
m = ± c2/ √(a2 + b2 + c2)
Note that two parallel lines have the same set direction cosines if we are provided with a line that doesn’t pass through the origin. The students need to create a new line from the origin parallel to the given line. Now you can find the direction cosines of the new line, which will be the same as the original line.
The relation between the direction cosines is that the sum of squares of the direction cosines must be equal to one. For the line having direction cosines x, y, and z, the relation is
x2 + y2 + z2 = 1
Direction Cosine For a Line Between Two Points
Suppose there is a line RS such that R(a1, b1, c1) and S(a2, b2, c2). The direction cosine between these points is formulated as:
x = a2 – a1/ RS
y = b2 – b1/ RS
z = c2 – c1/ RS
Here RS = √ (a2 – a1)2 + (b2 – b1)2 + (c2 – c1)2
Line Equations
Equation Of a Line in 3D Space
The equation of a line can be determined in two ways
- If the direction of the line along with the point through which it passes is given.
- The line passed from two known points.
- Equation of a line through a point parallel to a given vector
Vector Form:
The equation of the line in vector form is given by the formula
r= x + λy
Where r = the vector of the arbitrary position of a point on the line
x = the position vector of any point A
y = the vector parallel to x
λ = Any real number
Cartesian Form:
The equation of the line in vector form is given by the formula
(x – x1)/ a = (y – y1)/ b = (z – z1)/ c
Where (x1, y1, z1) are the coordinates of the known point and (a, b, c) are the direction cosines of the line.
- Equation of a line through a point parallel to a given vector
Vector Form:
The equation of the line in vector form is given by the formula
r= x + λ(y – x)
Where r = the vector of the arbitrary position of a point on the line
x = the position vector of any point A
y = the vector parallel to x
λ = Any real number
Cartesian Form:
The equation of the line in vector form is given by the formula
(x – x1)/ (x2 – x1) = (y – y1)/ (y2 – y1) = (z – z1)/(z2 – z1)
The above equation is also known as the standard cartesian equation of a line where (x1, y1, z1) and (x2, y2, z2) are the coordinates of the known points.
Angle Between Two Lines
Consider two lines L1 and L2 having direction cosines as (x1, y1, z1) and (x2, y2, z2) respectively. The angle formed between the two line is determined by using the formula:
Cos θ = (x1x2 + y1y2 + z1z2) / (√x12 + y12 + z12 . √x22 + y22 + z22)
Here x1i + y1j + z1k and x2i + y2j + z2k are two vectors intersecting at a point A.
Special Cases:
- When the lines are perpendicular → x1x2 + y1y2 + z1z2 = 0
- When the lines are parallel to each other → x1/x2 = y1/y2 = z1/z2
Shortest Distance Between Two Lines
The shortest distance between lines L1 and L2 given by the formulas r= x1 + λy1 and r= x2 + y2 is formulated as:
D = Modulus [ (y1 ⨯ y2) . (x1⨯ x2) / |(y1 ⨯ y2)| ]
Plane Equations
Equation of a Plane in 3D
The equation of a plane in 3D geometry is given by the matrix:
A =
This matrix is equal to zero. A = 0.
The vector form of a plane in the normal form is given as:
- r. n = d. Here vector r is the position vector of a point on the plane, ‘n’ is the normal and d is the length covered by the normal from the origin to the plane.
The cartesian form of a plane is lx + my + nz + d = 0 for any point P(x,y,z) having direction cosines l, m, and n.
Intercept Formula of a Plane
The formula of a plane passing through the points (a,0,0), (0, b, 0), and (0,0,c) is given as:
xa + yb + zc = 1
Coplanarity of Two Lines
Two lines are said to lie on the same plane if line 1 = x1 +λy1 and line 2 = x2 +y2 are such that (x1 – x2). (y1 ⨯ y2) = 0.
Angle Between Two Planes
The angle between two planes A and B having normal form equation as r. n1 = d1 and r. n2 = d2 is given by the formula:
Vector form: Cos θ = |n1.n2||n1|.|n2|
Cartesian Form: Cos θ = Modulus [ (x1x2 + y1y2 + z1z2) / (√x12 + y12 + z12 . √x22 + y22 + z22)]. Here (x1, y1, z1) and (x2, y2, z2) are the direction ratios of the planes.
Equation Of a Plane Through The Intersection of Two Planes
Vector Equation
Let the two planes be X and Y with the vector formula r. n1 = d1 and r. n2 = d2 respectively. The vector equation for plane passing through the intersection of two planes is:
- r. (n1 + λn2) = d1 +λd2
Cartesian Equation
The cartesian equation of a plane which passes through the intersection of two given planes P1x + Q1y + R1z + S1 = 0 and P2x + Q2y + R2z + S2 = 0 is formulated as (P1x + Q1y + R1z + S1) + λ(P2x + Q2y + R2z + S2) = 0.
Distance Between a Point And Plane
The distance between a point X (k, l, m) from a plane Ax + By + Cz = D is given as
D = Modulus [(Ak + Bl + Cm – D) / (√A2 + B2 + C2)]
Conclusion
3D geometry and vectors is a formula-based chapter. A student may find this hard due to the plethora of formulas and Maths concepts. Once a student gets an idea about how the formulas were derived, they can easily understand the ideas of this chapter. There are short as well as long formulas in this chapter. All of them are necessary for CBSE boards and JEE 2022 and other olympiad exams 2021-22. Long answer-type questions will appear in this chapter, so a student must be well prepared with all the topics and formulas. Every question in the NCERT Solutions for class 12 must be solved twice to increase the speed of the final examination. A student must seek help from teachers or NCERT solutions for 3D geometry for answers to the problems. They can also look for problems in previous year’s question papers and solve them to achieve a better score.
