Point and Line in 3D: Consider the position of a cricket ball at various points in time or the position of a butterfly as it flies from one flower to another at various points during its flight. Suppose we want to find the lowest tip of a Christmas star hanging from the ceiling or the central point of a hanging light in a room.

We require not two but three numbers representing the perpendicular distances of the point from three mutually perpendicular planes. The three distances are referred to as the point’s coordinates with reference to the three coordinate planes. In this article, let us learn about points and lines in a three-dimensional space.

Point and Line in 3D: Coordinate Axes and Planes

Consider three planes intersecting at \(O\) and being mutually perpendicular to each other. These three planes intersect along the lines \(X’OX,Y’OY,\) and \(Z’OZ\) are known as the \(x,y,\) and \(z\)-axes.
We can see that these lines are perpendicular to one another. The rectangular coordinate system is made up of these lines. The three coordinate planes are \(XOY,YOZ\) and \(ZOX,\) also known as the \(XY\)-plane, \(YZ\)-plane, and \(ZX\)-plane, respectively. We consider the \(XOY\) plane to be the plane of the paper and the line \(Z’OZ\) to be perpendicular to the \(XOY\) plane.
If we consider the plane of the paper to be horizontal, the line \(Z’OZ\) will be vertical. Distances measured from the \(XY\)-plane upwards in the direction of \(OZ\) are considered positive, while distances measured downwards in the direction of \(OZ’\) are considered negative. Similarly, we can say that the distance measured to the right of the \(ZX\)-plane along \(OY\) is positive, the distance measured to the left of the \(ZX\)-plane and along \(OY’\) is negative, the distance measured in front of the \(YZ\)-plane along \(OX\) is positive, and the distance measured to the back of it along \(OX’\) is negative. The coordinate system’s origin is designated as point \(O.\) These three planes divide the space into eight sections called octants. These octants could be named as \(XOYZ,X’OYZ,X’OY’Z,XOY’Z,XOYZ’,X’OYZ’,X’OY’Z’\) and \(XOY’Z’\) and denoted by \(I,II,III,….,VIII,\) respectively.

Coordinates of a Point in Space

The coordinates of the origin are written as \(\left( {0,0,0} \right).\) The coordinates of any point on the \(x\)-axis will be as \(\left( {x,0,0} \right),\) and the coordinates of any point in the \(YZ\)-plane will be as \(\left( {0,y,z} \right).\)

The sign of the coordinates of a point determines the octant where the point lies. The table below shows the signs of coordinates in the eight octants.

Octants Coordinates	I	II	III	IV	V	VI	VII	VIII
\(x\)	\( + \)	\( – \)	\( – \)	\( + \)	\( + \)	\( – \)	\( – \)	\( + \)
\(y\)	\( + \)	\( + \)	\( – \)	\( – \)	\( + \)	\( + \)	\( – \)	\( – \)
\(z\)	\( + \)	\( + \)	\( + \)	\( + \)	\( – \)	\( – \)	\( – \)	\( – \)

Distance Between Two Points

The distance between two points \(\left( {{x_1},{y_1},{z_1}} \right)\) and \(\left( {{x_2},{y_2},{z_2}} \right)\) is the shortest distance between them. It is equal to the square root of the summation of the square of the difference of the \(x\)- coordinates, the \(y\)-coordinates, and the \(z\)-coordinates of the two given points. The formula given below can be used to calculate the distance between two points.
\(D = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2} + {{\left( {{z_2} – {z_1}} \right)}^2}} \)
In particular, if \({x_1} = {y_1} = {z_1} = 0,\) i.e., point \(P\) is origin \(O,\) then \(OQ = \sqrt {{x_2}^2 + {y_2}^2 + {z_2}^2} \)
which gives the distance between the origin \(O\) and any point \(Q\left( {{x_2},{y_2},{z_2}} \right).\)

Direction Cosines of a Line

If a directed line \(L\) passes through the origin and forms angles \(\alpha ,\beta ,\) and \(\gamma \) with the \(x,y,\) and \(z\)-axes, respectively, these angles are called direction angles. The cosines of these angles named as \(\cos \alpha ,\cos \beta \) and \(\cos \gamma ,\) are called direction cosines of the directed line \(L.\)
Because a given line in space can be extended in two opposing directions, it has two sets of direction cosines. For a set of direction cosines of a line to be unique, we must first consider the given line to be a directed line. These distinct direction cosines are denoted by the letters \(l,m,\) and \(n.\)
If the given line does not pass through the origin \(O,\) we draw a line parallel to the given line that passes through the \(O.\) Now, find the direction cosines of the directed line from the origin because parallel lines have the same direction cosines.

Direction Ratios of a Line

The direction ratios of a line are any three numbers proportional to the direction cosines of a line. If \(l,m,n\) are direction cosines and \(a,b,c\) are line direction ratios, then \(a = \lambda l,b = \lambda m,\) and \(c = \lambda n\) for any non-zero \(R.\)
For a line, there are infinitely many sets of direction ratios. Two sets of direction ratios of the same line are proportional.
Consider a line with direction cosines \(l,m,n,\) then the relation between direction cosines is \({l^2} + {m^2} + {n^2} = 1\)
Since only one line can pass through two given points, we determine the direction cosines of the line passing through \(P\left( {{x_1},{y_1},{z_1}} \right)\) and \(Q\left( {{x_2},{y_2},{z_2}} \right)\) using \(PQ = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2} + {{\left( {{z_2} – {z_1}} \right)}^2}} \)

Equation of a Line in 3D

A line in \(3D\) can be represented by the vector and cartesian form of equations. A line is unique if one of the conditions is satisfied.
(i) The line passes through a given point and has a given direction
(ii) The line passes through two given points

Line Through a Point and Parallel to \(\overrightarrow b \)

Let \(\overrightarrow a \) be the position vector of the point A with respect to the origin \(O\) of the rectangular coordinate system.
Let
\(l\) be the line that passes through point \(A\) and is parallel to a given vector \(\overrightarrow b \)
\(\overrightarrow r \) be the position vector of an arbitrary point \(P\) that lies on the line \(\therefore \vec r = \vec a + \lambda \vec b\)
Cartesian equation of the line.
\(\frac{{\left( {x – {x_1}} \right)}}{l} = \frac{{\left( {y – {y_1}} \right)}}{m} = \frac{{\left( {z – {z_1}} \right)}}{n}\)
Where \(l,m\) and \(n\) are non-zero numbers.

Line Passing Through Two Given Points

Let \(\overrightarrow a \) and \(\overrightarrow b \) be the position vectors of two points \(A\left( {{x_1},{y_1},{z_1}} \right)\) and \(B\left( {{x_2},{y_2},{z_2}} \right)\) respectively that are lying on a line.
Let \(\overrightarrow r \) be the position vector of an arbitrary point \(P\left( {x,y,z} \right),\) then \(P\) is a point on the line if and only if \(\overrightarrow {AP} = \vec r – \vec a\) and \(\overrightarrow {AB} = \vec b – \vec a\) are collinear vectors.
\(\therefore \vec r = \vec a + \lambda \left( {\vec b – \vec a} \right),\lambda \in R\)
This is the vector form of the equation of the line.
The equation of the line passing through \(\left( {{x_1},{y_1},{z_1}} \right)\) and \(\left( {{x_2},{y_2},{z_2}} \right)\) in Cartesian form is given by \(\frac{{x – {x_1}}}{{{x_2} – {x_1}}} = \frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \frac{{z – {z_1}}}{{{z_2} – {z_1}}}\)

Solved Examples

Q.1. In the figure, if \(P\) is \(\left( {2,4,5} \right),\) find the coordinates of \(F.\)

Ans: For point \(F,\) the distance measured along \(OY\) is zero. Therefore, the coordinates of \(F\) are \(\left( {2,0,5} \right).\)

Q.2. Find the octant where the points \(\left( { – 3,1,2} \right)\) and \(\left( { – 3,1, – 2} \right)\) lie.
Ans: The point \(\left( { – 3,1,2} \right)\) lies in the second octant and the point \(\left( { – 3,1, – 2} \right)\) lies in octant \(VI.\)

Q.3. Find the distance between the points \(P\left( {1, – 3,4} \right)\) and \(Q\left( { – 4,1,2} \right).\)
Ans: The distance \(PQ\) between the points \(P\left( {1, – 3,4} \right)\) and \(Q\left( { – 4,1,2} \right)\) is \(PQ\)

This can be found using the distance formula
\(D = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2} + {{\left( {{z_2} – {z_1}} \right)}^2}} \)
\( = \sqrt {{{\left( { – 4 – 1} \right)}^2} + {{\left( {1 – – 3} \right)}^2} + {{\left( {2 – 4} \right)}^2}} \)
\( = \sqrt {{{\left( { – 5} \right)}^2} + {4^2} + {{\left( { – 2} \right)}^2}} \)
\( = \sqrt {\left( {25 + 16 + 4} \right)} \)
\( = \sqrt {45} = 6.7{\text{units}}\)

Q.4. Find the equation of the line passing through \(P = \left( {3, – 1,2} \right)\) and \(Q = \left( { – 3,0,1} \right).\)
Ans: First, we need to find the direction vector. To get the direction vector, we need to subtract the position vectors of the two points.

\(\vec d = \overrightarrow {PQ} = \left( {{d_1},{d_2},{d_3}} \right) = \left( {\left( {{x_2} – {x_1}} \right),\left( {{y_2} – {y_1}} \right),\left( {{z_2} – {z_1}} \right)}) \right.\)
\( = \left( {\left( { – 3 – 3} \right),\left( {1 – 0} \right),\left( {1 – 2} \right)} \right)\)
\( = \left( { – 6,1, – 1} \right)\)

Now, the equation of the line can be obtained as
\(\frac{{\left( {x – {x_1}} \right)}}{{{d_1}}} = \frac{{\left( {y – {y_1}} \right)}}{{{d_2}}} = \frac{{\left( {z – {z_1}} \right)}}{{{d_3}}}\)

Considering \(P\) as \(\left( {{x_1},{y_1},{z_1}} \right),\) the equation of the line is
\(\frac{{x – 3}}{{ – 6}} = \frac{{y – \left( { – 1} \right)}}{1} = \frac{{z – 2}}{{ – 1}}\)
\( = \frac{{3 – x}}{6} = y + 1 = 2 – z\)

Q.5. Find the vector and Cartesian equations of the line through the point \(\left( {5,2, – 4} \right)\) and which is parallel to the vector \(3\hat \imath + 2\hat j – 8\hat k.\)
Ans: We \(a = 5\hat \imath + 2\hat \jmath – 4\hat k\) and \(b = 3\hat \imath + 2\hat \jmath – 8\hat k.\)

The vector equation of the line is
\(\vec r = 5\hat \imath + 2\hat \jmath – 4\hat k + \lambda \left( {3\hat \imath + 2\hat \jmath – 8\hat k} \right)\)

Now, \(\vec r\) is the position vector of \(P\left( {x,y,z} \right)\) on the line.
\(\therefore x\hat \imath + y\hat \jmath + z\hat k = 5\hat \imath + 2\hat \jmath – 4\hat k + \lambda \left( {3\hat \imath + 2\hat \jmath – 8\hat k} \right)\)
\(\left( {5 + 3\lambda } \right)\hat i + \left( {2 + 2\lambda } \right)\hat j – \left( { – 4 – 8\lambda } \right)\hat k\)

Eliminating \(\lambda ,\) we get
\(\frac{{x – 5}}{3} = \frac{{y – 2}}{2} = \frac{{z + 4}}{{ – 8}}\)

This is the equation of the line in Cartesian form.

Summary

Direction cosines of a line are the cosines of the angles made by the line with the positive directions of the coordinate axes. Direction ratios of a line are the numbers that are proportional to the direction cosines of a line. We can find the vector equation of a line that passes through the given point whose position vector is \(\overrightarrow a \) and parallel to a given vector \(\overrightarrow b .\) Equation of a line through a point \(\left( {{x_1},{y_1},{z_1}} \right),\) and having direction cosines \(l,m,n\) can also be determined. The vector equation and cartesian equation of a line that passes through two points can also be determined.

Frequently Asked Questions (FAQs)

Students must be having many questions regarding Point and Line in 3D. Here the commonly asked questions are compiled below:

Q.1. How do you find the points on a line in 3D?
Ans: The equation of the line with the direction vector \(\vec d = \left( {l,m,n} \right)\) and that passes through the point \(P\left( {{x_1},{y_1},{z_1}} \right)\) is found by the formula \(\frac{{\left( {x – {x_1}} \right)}}{l} = \frac{{\left( {y – {y_1}} \right)}}{m} = \frac{{\left( {z – {z_1}} \right)}}{n}\)

Where \(l,m\) and \(n\) are non-zero numbers. To check whether any point lies on the line, we have to check whether the given point satisfies the equation of the line.

Q.2. What is a line in 3D?
Ans: A line is a continuous and infinite succession of points in one dimension, whereas a plane is an ideal object with only two dimensions and infinite points and lines. If we look at a room, a wall will be a graph model of a plane, while where two walls intersect will give us the idea of a line.

Q.3. How do you find the distance between two points on a 3D plane?
Ans: The distance between two points \(\left( {{x_1},{y_1},{z_1}} \right)\) and \(\left( {{x_2},{y_2},{z_2}} \right)\) is the shortest distance between them. It is equal to the square root of the summation of the square of the difference of the \(x\)- coordinates, the \(y\)-coordinates, and the \(z\)-coordinates of the two given points.

The formula for the distance between two points is \(D = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2} + {{\left( {{z_2} – {z_1}} \right)}^2}} \)

Q.4. How do you find the direction cosines of a line passing through two points?
Ans: Since one and only one line passes through two given points, we can determine the direction cosines of a line passing through the given points \(P\left( {{x_1},{y_1},{z_1}} \right)\) and \(Q\left( {{x_2},{y_2},{z_2}} \right)\) as \(PQ = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2} + {{\left( {{z_2} – {z_1}} \right)}^2}} \)

Q.5. What is the equation of a line between two points in vector and cartesian form?
Ans: Let \(\overrightarrow a \) and \(\overrightarrow b \) be the position vectors of two points \(A\left( {{x_1},{y_1},{z_1}} \right)\) and \(B\left( {{x_2},{y_2},{z_2}} \right)\) respectively that are lying on a line. The equation of the line in Cartesian form is given by \(\frac{{x – {x_1}}}{{{x_2} – {x_1}}} = \frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \frac{{z – {z_1}}}{{{z_2} – {z_1}}}\)

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