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Three Dimensional Geometry For Class 12

Three Dimensional Geometry for class 12 covers important topics such as direction cosine and direction ratios of a line joining two points. The notes are prepared as per the latest CBSE syllabus (2022-2023) and NCERT curriculum. Students can revise Maths Chapter 12 (Introduction to three-dimensional geometry) with the help of notes formulated as per the latest exam pattern. Also, we will discuss here, the equation of lines and planes in space under different conditions, the angle between line and plane, between two lines etc. You need to practice the questions to understand the topic better and based on the formulas as well.

To understand the different types of shapes and figures, this topic has been introduced in Maths. In the real world, almost all the objects are in a three-dimensional shape.  For example, there are many objects at home such as a table, chair, bed, kitchen utensils, etc. which have 3D geometry. In our primary classes, we have learned the basics of three-dimension geometry, but in the 12th standard, we will learn the advanced version of it.

Three Dimensional Geometry Class 12 Notes

Let us discuss some important formulas of three-dimensional geometry, based on which we will solve problems.

Direction Cosine and Direction Ratios of a line

Consider a line L passing through origin makes an angle of α, β, γ with x, y, and z-axes respectively, then the cosine of these angles is the direction cosine of the directed line L.

Any three numbers which are proportional to the direction cosines of a line are called the direction ratios of the line. Consider the direction cosine of line L be l, m, n and direction ratio a, b, c then a = λl, b = λm, c = λn, for non-zero λ ∈ R.

\(\begin{array}{l}\frac{l}{a} = \frac{m}{b} = \frac{n}{c} = k\end{array} \)

Then the direction cosine are:

\(\begin{array}{l}l = \pm \frac{a}{a^{2} + b^{2} + c^{2}}, m = \pm \frac{b}{a^{2} + b^{2} + c^{2}}, n = \pm \frac{c}{a^{2} + b^{2} + c^{2}}\end{array} \)

Note: If the given line in space does not pass through the origin, then, in order to find its direction cosines, we draw a line through the origin and parallel to the given line. Now take one of the directed lines from the origin and find its direction cosines as two parallel line have same set of direction cosines.

Relation between the direction cosines of a line

For a given line RS with direction cosines l, m, n, the relation between the direction cosines of a line RS is given by;

l2 + m2 + n2 = 1

Direction cosines of a line passing through two points

The direction cosines of the line segment joining the points P(x1, y1, z1) and Q(x2, y2, z2) are:

x2-x1/PQ, y2-y1/PQ, z2-z1/PQ

where PQ = √[(x2-x1)2+(y2-y1)2+(z2-z1)2]

Important Points to remember

  • The shortest distance between two skew lines is the line segment perpendicular to both lines.
  • If l, m, n are the direction cosines, and a, b, c are the direction ratios of the line, then the direction cosines are given by: 
    \(\begin{array}{l}l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
  • Skew lines are the lines that lie in different planes. They are neither parallel and nor intersecting.
  • \(\begin{array}{l} \text{The vector equation of the line that passes through two points, in which whose position } \\ \text{ vector is  } \vec{a} \text{ and parallel to the vector } \vec{b} \text{ is given by }\vec{r}= \vec{a}+\lambda \vec{b}\end{array} \)
  • \(\begin{array}{l} \text{The vector equation of the line that passes through two points, in which whose position } \\ \text{ vector is }\vec{a} \text { and } \vec{b} \text{ is given by } \vec{r}= \vec{a}+\lambda ( \vec{b}- \vec{a})\end{array} \)
  • The cartesian equation of a line that passes through the points (x1, y1, z1) and (x2, y2, z2) is given by
    \(\begin{array}{l}\frac{x-x_{1}}{x_{2}-x_{1}}= \frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\end{array} \)
  • The equation of the plane that passes the three non-collinear points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is
    \(\begin{array}{l}\begin{vmatrix} x-x_{1} & y-y_{1} & z-z_{1}\\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1}\\ x_{3}-x_{1} &y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}=0\end{array} \)
  • \(\begin{array}{l} \text{The vector equation of a plane which contains three non collinear points having the } \\ \text{ position vectors } \vec{a}, \vec{b},\vec{c} \text{ is given by } (\vec{r}-\vec{a}).[(\vec{b}-\vec{a})\times (\vec{c}-\vec{a})]=0\end{array} \)
  • The equation of the plane which cuts the coordinate axes at (a, 0, 0), (0, b, 0) and (0, 0, C) is given by
    \(\begin{array}{l}\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\end{array} \)
  • \(\begin{array}{l} \text{Two lines } \vec{r}= \vec{a_{1}}+\lambda\vec{b_{1}} \text { and }\vec{r}= \vec{a_{2}}+\mu\vec{b_{2}} \text {are said to be coplanar if } \\ (\vec{a_{2}}-\vec{a_{1}}).(\vec{b_{1}}\times \vec{b_{2}})=0\end{array} \)
  • The distance from the point (x1, y1, z1) to the plane Ax+By+Cz+D= 0 is given by:
    \(\begin{array}{l}\left | \frac{Ax_{1}+By_{1}+Cz_{1}+D}{\sqrt{A^{2}+B^{2}+C^{2}}} \right |\end{array} \)
  • \(\begin{array}{l} \text {If θ is the angle between the two planes in the vector form, } \vec{r}. \vec{n_{1}}= d_{1} \\ \text{ and } \vec{r}. \vec{n_{2}}=d_{2} \text {, then } \theta = cos^{-1}\frac{|\vec{n_{1}}.\vec{n_{2}}|}{|\vec{n_{1}}||\vec{n_{2}}|}\end{array} \)
  • The cartesian equation of a plane which passes through the intersection of two given planes A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0 is (A1x + B1y + C1z + D1 ) + λ(A2x + B2y + C2z + D2 ) = 0.
  • \(\begin{array}{l} \text{The distance of a point whose position vector is given as }\vec{a} \text { from the plane } \\  \vec{r}.\widehat{n}=d \text { is } |d-\vec{a}.\widehat{n}|\end{array} \)
  • \(\begin{array}{l}\text{In vector form,  the equation of a plane which is at a distance “d” from the point origin, } \\ \text{ and } \widehat{n} \text{ is the normal unit vector to the plane that passes through the origin is } \\ \text{given by } \vec{r}.\widehat{n}=d\end{array} \)
  • The equation of a plane which is at a distance of “d” from the origin, and the direction cosines of the normal to the plane are l, m, n is given by lx + my + nz = d.  
  • \(\begin{array}{l}\text{The equation of a plane that passes through a point, in which whose position vector is } \\ \vec{a} \text{ and perpendicular to the }\vec{N} \text{ is  } (\vec{r}-\vec{a}).\vec{N}=0\end{array} \)
  •  The equation of a plane perpendicular to a given line with the direction ratios A, B, C and passing through a given point (x1 , y1 , z1 ) is given by A (x – x1 ) + B (y – y1 ) + C (z – z1 ) = 0
  • The angle “θ” between two planes A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0 is given by
    \(\begin{array}{l}cos\theta =\left | \frac{A_{1}A_{2}+B_{1}B_{2}+C_{1}C_{2}}{\sqrt{A_{1}^{2}+B_{1}^{2}+C_{1}^{2}}\sqrt{A_{2}^{2}+B_{2}^{2}+C_{2}^{2}}} \right |\end{array} \)
  • The direction cosines of two lines are l1 , m1 , n1 and l2 , m2 , n2 and “θ” is the acute angle between the two lines; then cosθ = | l1 l 2 + m1 m2 + n1 n2|.

More Articles for Class 12

Solved Examples on Class 12 Three Dimensional Geometry

Question 1: 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.

Here let α = 90°, β = 135° and γ = 45°

As we know,

l = cos α, m = cos β and n = cos γ

Thus, direction cosines are:

l = cos 90° = 0

m = cos 135°= cos (180° – 45°) = -cos 45° = -1/√2

n = cos 45° = 1/√2

Therefore, the direction cosines of the line are 0, -1/√2, 1/√2.

Question 2: Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.

Solution:

If the direction ratios of two lines segments are proportional, then the lines are collinear.

Let the given points are: A(2, 3, 4), B(−1, −2, 1) and C(5, 8, 7)

Direction ratio of the line joining the points A (2, 3, 4) and B (−1, −2, 1), are:

((−1−2), (−2−3), (1−4)) = (−3, −5, −3)

Where, a1 = -3, b1 = -5, c1 = -3

Direction ratio of the line joining the points B (−1, −2, 1) and C (5, 8, 7) are:

[(5− (−1)), (8− (−2)), (7−1)] = (6, 10, 6)

Where, a2 = 6, b2 = 10 and c2 =6

Since, it is clear that the direction ratios of AB and BC are of same proportions, hence,

a2/a1 = 6/-3 = -2

b2/b1 = 10/-5 = -2

c2/c1 = 6/-3 = -2

Therefore, the points A, B and C are collinear.

Practice Problems on Class 12 Three Dimensional Geometry

Solve the following problems on three-dimensional geometry:

  1. Determine the equation of the plane that passes through the points (-4, -2, 3), (6, 4, -5) and  (1, 1, -1)
  2. Calculate the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin.
  3. \(\begin{array}{l}\text{Prove that the lines }\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1} \text{ and }\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text{ are perpendicular to each other.}\end{array} \)
  4. Prove that the lines that pass through the points (2, 3, 4) (4, 7, 8) are parallel to the line that passes through the points (1, 2, 5), (-1, -2, 1).
  5. Calculate the vector equation of the line that passes through the points (3, 4, 6) and (-1, 0, 2).

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