Three Dimensional Geometry

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In real-world, almost all the objects are in a three-dimensional shape. To understand the different types of shapes and figures, this topic has been introduced in Mathematics. For example, there are many objects at home such as a table, chair, bed, kitchen utensils, etc. which have three-dimensional shape.
Three Dimensional Geometry (3D) covers some important topics such as direction cosine, direction ratios of a line joining two points and also equation of lines and planes.

Coordinate System

We know that, cartesian plane is determined by two perpendicular number lines called x-axis and y-axis. These axes together with their point of intersection (origin O) gives a two-dimensional coordinate system.
Now, we introduce a third dimension to this model, then we will get three-dimensional coordinate system.

We can construct three-dimensional coordinate system by passing a z-axis perpendicualtr to both x-axis and y-axis at the origin O. Taken as pairs, then the axes determine three mutually perpendicualr coordinate planes: xy-plane, yz-plane and xz-plane.
In this system, a point P in space is deermined by an ordered triple \( (x,y,z ) \) where \( x,y,z\) are as follows

  1. \( x=\) directed distance from yz-plane to P
  2. \( y=\) directed distance from xz-plane to P
  3. \( z=\) directed distance from xy-plane to P

Direction Cosine

Let L is given line where P is any arbitrary point on L, where \(OP=r\) and L makes angles \( \alpha , \beta, \gamma \) with x, y and z-axes, respectively, called direction angles, then cosine of these angles, namely, \( \cos \alpha , \cos \beta, \cos \gamma \) are called direction cosines of line L, it is written as

  1. \(\cos \alpha=l\)
  2. \(\cos \beta=m\)
  3. \(\cos \gamma=n\)

In the figure 1 given above, \(PA \perp \text{x-axis}\) , so the triangle OAP is right angled triangle whose \(OP=r\), \(OA=x\), right angle is at A, and angle at O is \(\alpha\)
Therefore
\( \cos \alpha=\frac{x}{r}\)
or \( x=r \cos \alpha\)
or \( x=lr\)
Similarly, in the figure 2 given above, \(PB \perp \text{y-axis}\) , so the triangle OBP is right angled triangle whose \(OP=r\), \(OB=y\), right angle is at B, and angle at O is \(\beta\)
Therefore
\( \cos \beta=\frac{y}{r}\)
or \( y=r \cos \beta\)
or \( y=mr\)
Similarly, in the figure 3 given above, \(PC \perp \text{z-axis}\) , so the triangle OCP is right angled triangle whose \(OP=r\), \(OC=z\), right angle is at C, and angle at O is \(\gamma\)
Therefore
\( \cos \gamma=\frac{z}{r}\)
or \( z=r \cos \gamma\)
or \( z=nr\)
So,
\( l^2+m^2+n^2=1\)

Prove that \( l^2+m^2+n^2=1\)

Let \(P(x,y,z)\) be any arbitrary point on the directed line OP, then we draw
\( PQ \perp \text{xy-plane}\)
Then, from the figure 1, we get that
\( OQ^2=x^2+y^2\)
Again, from figure 2, from right angled triangle OPQ, , where \(OP=r\), then we get that
\( OP^2=OQ^2+PQ^2\)
or \( r^2=x^2+y^2+z^2\)
or \( x^2+y^2+z^2=r^2\)
Now, we know that
\(x=lr,y=mr,z=nr\)
Therefore
\(x^2+y^2+z^2=r^2\)
or \( (lr)^2+(mr)^2+(nr)^2=r^2\)
or \(l^2+m^2+n^2=1\)

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.
Any three numbers which are proportional to the direction cosines of a line are called the direction ratios of the line.
If \(l, m, n\) are direction cosines and \( a, b, c \) are direction ratios of a line, then
\( \frac{a }{l} = \frac{b }{m} =\frac{c }{n} =\lambda \) , for any nonzero \( \lambda \in R \)

Straight Line

To find the equation of a line in a two-dimensional plane, we need to know a point that the line passes through as well as the slope.
Similarly
in three-dimensional space, we can obtain the equation of a line if we know a point that the line passes through as well as the direction vector (direction cosine), which designates the direction of the line. The formula is as follows:

  1. The equation of a line with direction \( \vec{d}=(l,m,n) \) and passes through \( (x_1,y_1,z_1) \) is
    \( \frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n} \)
    Proof
    Consider a line \(l\) which passes through a point \( A=(x_1,y_1,z_1) \) and has direction \( \vec{d}=(l,m,n) \).
    Let \( P=(x,y,z)\) be an arbitrary point on the line \(l\)
    Then the vector \( \vec{PA} \) will be parallel to \( \vec{d} \).
    Hence we have
    \( \overrightarrow{PA} = \lambda \vec{d} \)
    or\( (x-x_1, y-y_1, z-z_1 ) =\lambda (l,m,n) \)
    or\( \frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n} \).
    This is the required equation of the straight line \(l\)
  2. The equation of a line through \( (x_1,y_1,z_1) \) is \( (x_1,y_1,z_1) \) is
    \( \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} \)
    We know that the equation of a straight line \(l\) which passes through the point \( A=(x_1,y_1,z_1) \) be
    \( \frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n} \) (i)
    Also, Consider that the line \(l\) passes through the point \( B=(x_2,y_2,z_2) \), then its equation is
    \( \frac{x-x_2}{l}=\frac{y-y_2}{m}=\frac{z-z_2}{n} \) (ii)
    Using the proportionality between (i) and (ii), the equation of the line \(l\) passing through \( (x_1,y_1,z_1) \) and \( (x_2,y_2,z_2) \) is
    \( \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} \)

The relationship between two different lines in a three-dimensional space is always one of the three: they can be parallel, skew, or intersecting at one point. If the direction vectors of the lines are parallel, then the lines are also parallel (provided that they are not identical). If the lines do not meet and their direction vectors are not parallel, then they are skew. If the lines meet and their direction vectors are not parallel, then the lines meet at a single point. As we can see, comparing the direction vectors usually gives useful information concerning two lines.

Example 1

  1. Find the equation of a line with direction vector \( \vec{d}=(1,2,3) \) that passes through the point \( P=(-1,0,1) \)
  2. Find the equation of the line that passes through the points \( P=(3,-1,2)\) and \( Q=(-3,0,1)\)
  3. Find the equation of the line that passes through the points \( P=(1,1,1)\) and \( Q=(-1,1,3)\)
  4. Find the relationship of the lines: \( \frac{x-2}{2}=y=\frac{z+1}{-3} \text{ and } \frac{x+7}{-6}=-\frac{y}{3}=\frac{z+1}{9}\)
    Solution
    The direction vectors of the two lines are
    \( \vec{d_1}=(2,1,-3) \) and \( \vec{d_2}=(-6,-3,9)\)
    Since
    \( -3\vec{d_1}=\vec{d_2}\)
    The two direction vectors are parallel.
    This implies that the two lines are either identical or parallel.
    A point that the first line passes through is (2,0,-1)
    Since this point does not satisfy the equation of the second line, the second line does not pass through this point.
    Therefore, the two lines are parallel.

Angle Bttween Two Lines

Let the direction cosines of the two lines are \( \vec{a}=(l_1, m_1, n_1) \) and \( \vec{b}=(l_2, m_2, n_2) \) respectively.
Now, let \(\theta\) be the angle between the lines.
Then, the angle between them is:
\( \cos \theta = \frac{\vec{a}. \vec{b}}{|\vec{a}|. |\vec{b}|}\)
or \( \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2}\sqrt{l_2^2 + m_2^2 + n_2^2}} \)
or \( \cos \theta =l_1 l_2 + m_1 m_2 + n_1n_2 \)

Example

Find the angle between the pair of lines \( \frac{x + 3}{3} = \frac{y – 1}{5} = \frac{z + 3}{4} \) and \( \frac{x + 1}{1} = \frac{y – 4}{1} = \frac{z – 5}{2} \)

Let \( \frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1} \). and \( \frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2} \) are two lines with direction \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \) respectively, then the lines are co-planar if
\( \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} =0 \)

Shortest Distance

The shortest distance between two parallel lines helps to determining how far apart lines are.
This can be done by measuring the length of a line that is perpendicular to both of them.
Thus, the shortest distance between two straight lines is the minimum distance between any two points lying on the lines.

Distance between two Parallel Lines The distance is the perpendicular distance from any point on one line to the other line.
Distance between two Intersecting Lines The shortest distance between such lines is zero.
Distance between two Skew Lines The distance is equal to the length of the perpendicular between the lines.

NOTE

Let \( \frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1} \). and \( \frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2} \) are two lines with direction \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \) respectively, then the shortest distance between them is
\( ( x_2-x_1 ) l+ ( y_2-y_1 )m+( z_2-z_1)n \)
where \( l,m,n\) actual direction cosine of the SD line which are obtained by
\( ll_1+mm_1+nn_1=0 \)
\( ll_2+mm_2+nn_2=0 \)

Exercise 1

  1. Find the direction cosines of the line passing through the two points \((– 2, 4, – 5)\) and \((1, 2, 3)\).
  2. Show that the points \(A (2, 3, – 4), B (1, – 2, 3)\) and \(C (3, 8, – 11)\) are collinear.
  3. If a line makes angles 90, 135, 45 with the x, y and z-axes respectively, find its direction cosines.
  4. Show that the lines \( \frac{x – 5}{7}= \frac{y + 2}{-5}= \frac{z}{1}\)1 and \( \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other.
  5. Find the direction cosines of a line whose direction ratios are 2, -6, 3. (Answer: 7)
  6. Find the direction cosines of a line that makes equal angles with the coordinate axes. (Answer: 1/√3, 1/√3, 1/√3)
  7. Find the angles of triangle ABC whose vertices are A(-1, 3, 2), B (2, 3, 5) and C(3, 5, -2). (Answer: A = pi/2, B =cos-1(1/√3), C =cos-1(√2/5)
  8. Find the angles between the lines whose direction ratios are 3, 2, -6 and 1, 2, 2. ( Answer: cos-1(5/21)
  9. 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? (Answer: 60 degrees)
  10. 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.

General equation of plane

An equation of the first degree in x,y,z represents a plane.
This equation is given by
ax+by+cz+d=0
where a,b,c are constant and not all zero.

The condition that a,b,c are not all zero is equivalent to
\( a^2+b^2+c^2≠0 \).

Example

Explain the plain ax+by+cz+d=0 when

  • a=0
  • b=0
  • c=0
  • d=0
  • a=0,b=0
  • a=0,c=0
  • b=0,c=0.

Various form of equation of plane

  1. General equation of a plane is \( ax+by+cz+d = 0 \)
  2. The equation of the plane whose intercepts are a, b, c on the x, y, z axes respectively is\( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 ; (a, b, c \ne 0) \)
  3. Equation of the plane in Normal form is\( lx + my + nz = p \)where p is the length of the normal from the origin to the plane and \( (l, m, n) \) be the direction cosines of the normal.
  4. The equation to the plane passing through \( P(x_1, y_1, z_1) \) and having direction ratios (a, b, c) for its normal is \( a(x – x_1) + b(y – y_1) + c (z – z_1) = 0 \)
  5. The equation of the plane passing through three non-collinear points \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), and \( (x_3, y_3, z_3) \),is \( \begin{vmatrix} x-x_1 & y- y_1 & z- z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} = 0 \)
  6. The equation of plane through three points \( (x_1,y_1,z_1), (x_2,y_2,z_2), (x_3,y_3,z_3) \) is
    \( \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \)
  7. Equation of YZ plane is x = 0
  8. Equation of plane parallel to YZ plane is x = d.
  9. Equation of ZX plane is y = 0
  10. Equation of plane parallel to ZX plane is y = d.
  11. Equation of XY plane is z = 0
  12. Equation of plane parallel to XY plane is z = d.
  13. Four points namely \( A =(x_1, y_1, z_1), B= (x_2, y_2, z_2), C= (x_3, y_3, z_3) \text{and} D =(x_4, y_4, z_4) \) will be coplanar if one point lies on the plane passing through other three points.
  14. Three points namely \( A =(x_1, y_1, z_1), B= (x_2, y_2, z_2), C= (x_3, y_3, z_3) \) will be coplanar if
    \( \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix}= 0 \)
  15. Four points namely \( A =(x_1, y_1, z_1), B= (x_2, y_2, z_2), C= (x_3, y_3, z_3) \text{ and } D =(x_4, y_4, z_4) \) will be coplanar if
    \( \begin{vmatrix} x_1 & y_1 & z_1 & 1\\ x_2 & y_2 & z_2 & 1\\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1\end{vmatrix}= 0 \)

Exercise 2

  1. Find the intercepts of the plane 2x-4y+6z=15
  2. Find the equation of a plane which makes intercepts 2,-3,5 in coordinate axes.
  3. Find the equation of the plane containing the lines through the origin with direction cosines proportional to 2, 1, –2 and 5, 2, –3.
    Solution
    The equation of the plane through origin is
    ax + by + cz = 0 (i)
    Also, the plane (i) contains the lines with direction cosines proportional to 2, 1, –2 and 5, 2, –3.
    Therefore, the normal to the plane will be perpendicular to these lines. By using the condition of perpendicularity, we have,
    5a + 2b – 3c = 0 (ii)
    2a + b – 2c = 0 (iii)
    Solving there equations (ii) and (iii) by cross – multiplication, we get
    a = 1, b = 4 and c = 1
    Hence, the plane is x – 4y – z = 0

Examples 2

  1. Find the equation of the plane through the points P(1, 1, 1),Q(3, 1, 2) and R(–3, 5, –4).
    Solution
    The equation of the plane is
    \( \begin{vmatrix} x-1 & y-1 & z-1 \\ 3 & 1 & 2 \\ -3 & 5 & 4 \end{vmatrix} \)
    or x + y = 2
  2. Angle between the planes is defined as angle between normals of the planes drawn from any point to the planes.
    Angle between the planes \( a_1x + b_1y + c_1z + d_1 = 0\) and \( a_2x + b_2y + c_2z + d_2 = 0 \) is

    Note:
    1. \( a_1a_2+ b_1b_2+ c_1c_2=0 \), then the planes are perpendicular to each other.
    2. If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) then the planes are parallel to each other.
    3. A plane parallel to the plane \( ax + by + cz + d = 0 \text{ is .} ax + by + cz + k = 0 \)
  3. Find the angle between a pair of planes \(2x-y+z=6\) and \(x+y+2z=7\)
  4. Find the equation if plane through \((2,-3,4)\) and parallel to \(2x-6y-7z=6\)
  5. Find the equation to the plane through the point (–1, 3, 3) and perpendicular to the planes \(x + 2y + 3z = 5\) and \(3x + 3y + z = 9\).
    Solution
    The plane passing through the point (–1, 3, 2) is
    a(x + 1) +b (y – 3) + c(z – 2) = 0 (i)
    The plane (i) is perpendicular to the planes,x + 2y + 3z = 5 and 3x + 3y + z = 9.
    Then applying the condition of perpendicularity, we get
    a.1 + b.2 + c.3 = 0(ii)
    a.3 + b.3 + c.1 = 0(iii)
    From (ii) and (iii), by cross multiplication, we get
    a=-7, b=8, c=-3
    Substituting these values in (i), we get
    – 7k (x + 1) + 8k (y – 3) – 3k (z – 2) = 0
    or 7x – 8y + 3z + 25 = 0
    Which is required equation of the plane.
  6. Find the equation of the plane through the points (–1, 1, –1) and (6, 2, 1) normal to the plane 2x+y+z=5
    Solution
    Any equation of the plane through the point (–1, 1,–1) is
    a(x + 1) + b(y – 1) + c(z + 1) = 0(i)
    The plane (i) passes through (6, 2, 1), so
    7a + b + 2c = 0 (ii)
    Since the plane (i) is normal to the plane 2x + y + z = 5, we have
    2a + b + c = 0 (iii)
    Form (ii) and (iii) we get,
    a=1, b=3, c=-5 (iv)
    Hence the plane is
    x + 3y – 5z = 7
  7. Find the equation of the plane through the point (1, 1, 1) and perpendicular to the plane x – 2y + z = 2 and 4x + 3y– x + 1 = 0
    Solution
    Any plane through the point (1, 1, 1) is
    a(x – 1) + b(y – 1) + c(z – 1) = 0 (i)
    Since (i) is perpendicular to the planes x – 2y + z = 2 and 4x + 3y – z + 1 = 0, therefore
    a – 2b + c = 0 (ii)
    4a + 3b – c = 0 (iii)
    Solving these equations (ii) and (iii) by cross-multiplication, we get
    a = –1, b = 5 and c = 11
    Putting the values of a, b and c in (i), we get
    x – 5y – 11z + 15 = 0

Angle between a line and a plane

Angle between a line and a plane is defined as angle between the line and normal of the plane drawn from any point to the plane.
Angle between a line AB \( (a_1,b_1,c_1)\) and a plane P \( a_2x + b_2y +c_2z + d_2 = 0 \) is
\( cos(90-\theta) =\frac{a_1a_2+ b_1b_2+ c_1c_2}{\sqrt{ a_1^2 + b_1^2 + c_1^2}\sqrt{ a_2^2 + b_2^2 + c_2^2} } \)

Note:
  1. The line is parallel to the plane if \( a_1a_2+ b_1b_2+ c_1c_2=0 \).
  2. If \( l,m,n \) be the direction cosines of the normal to the plane the cosines which the plane makes with yz,zx and xy plane.
  3. The equation of a plane perpendicular to the x-axis is \( x=p\)
    or \( l=1,m=0,n=0 \)
  4. The equation of a plane parallel to the x-axis is \( my+nz=p \)

Examples

  1. Find the equation of the plane passing through the intersection of the planes \(x + y + z =5\) and \(2x + 3y + 4z – 5 = 0\) and passing through
    Solution:
    The plane through the intersection of the given two planes is
    \((x+y+z - 5) + k(2x + 3y + 4z - 5) = 0\)
    Since the plane (i) passes through the origin. So, from (i),
    \(k = - 1\)
    Hence, the plane is
    \(x + 2y + 3z = 0\)
  2. Find the equation of the plane which is perpendicular to the plane \(5x + 3y + 6z + 8 = 0\) and which contains the line of intersection of the planes \(x + 2y + 3z – 4 = 0\) and \(2x + y – 2 + 5 = 0\)
    Solution
    A plane passing through the intersection of the planes
    x+2y+3z– 4+k(2x+y–z+5) = 0 (i)
    or x(1 + 2k) + y(2 + k) + z(3 – k) + (5k – 4) = 0
    Since plane (i) perpendicular to the plane
    5x + 3y + 6z + 8 = 0
    So (1 + 2k) 5 + 3(2 + k) + 6(3 – k) = 0
    or k = –297
    Putting this value in (i), we get
    (x + 2y + 3z – 4) – 297 (2x + y – z + 5) = 0
    or 51x + 15y – 50z + 173 = 0

The length of perpendicular

The length of perpendicular from a point \( P(x_1,y_1,z_1) \) to the plane \( ax+by+cz+d=0\) is
\( d=\frac{ ax_1+by_1+cz_1+d }{\sqrt{a^2+b^2+c^2}} \)

Exercise 3

  1. Find the perpendicular distance from a point (1,2,3) to the plane x+y+z-3=0
  2. Find the distance between two parallel planes 2x + 3y + 6z + 1 = 0 and 4x + 6y + 12x + 65 = 0
    Solution
    Given parallel planes are
    2x + 3y + 6z + 1 = 0 (i) and
    4x + 6y + 12z + 65 = 0 (ii)
    Since the distan between two parallel planes is the distance of any point on the plane from the other plane.
    So, for the coordinates of a point on the plane (i), Putting y = z = 0 then we get,
    \( x = –\frac{1}{2} \)
    It implies that any point on the plane (i) is \( (\frac{1}{2},0,0) \)
    Required distance is the distance of the point \( (\frac{1}{2},0,0) \) from plane (ii) is given by
    \( d = \frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}=\frac{9}{2}\)

Angle Bisector of two planes

Angle Bisector of two planes \( a_1x + b_1y + c_1z + d_1 = 0 \) and \( a_2x + b_2y + c_2z + d_2 = 0 \) is
\( \frac{ a_1x + b_1y + c_1z + d_1}{\sqrt{ a_1^2 + b_1^2 + c_1^2}} = \pm \frac{ a_2x + b_2y + c_2z + d_2 }{\sqrt{ a_2^2 + b_2^2 + c_2^2} } \)

Question

Find the Value of r

A rectangle of dimention 16cm by 12cm are titned by two circles of 5 cm and r cm radius as given below. Find the value of r.
\((5+r)^2=(16-5-r)^2+(7-r)^2\) implies that \(r=3.4\)

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