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Equation of a Straight Line in Different Forms Straight lines constitute an important topic of the three dimensional geometry. The topic is a bit tricky but with a little hard work can fetch you some direct questions. A straight line can be represented in different forms and questions are often framed in the exam on these forms. We shall first discuss some of the forms of representing the straight line and then proceed to some related questions. What exactly is a straight line? In three dimensional geometry, a straight line is defined as the intersection of two planes. So general equation of straight line is stated as the equations of both plane together i.e. general equation of straight line is a_{1}x + b_{1}y + c_{1}z + d_{1} = 0, a_{2}x + b_{2}y + c_{2}z + d_{2} = 0 ……(1) So, equation (1) represents straight line which is obtained by intersection of two planes. View the video on straight lines Equation of Straight Line in Different Forms: Symmetrical Form:
Straight lines constitute an important topic of the three dimensional geometry. The topic is a bit tricky but with a little hard work can fetch you some direct questions. A straight line can be represented in different forms and questions are often framed in the exam on these forms. We shall first discuss some of the forms of representing the straight line and then proceed to some related questions.
What exactly is a straight line?
In three dimensional geometry, a straight line is defined as the intersection of two planes. So general equation of straight line is stated as the equations of both plane together i.e. general equation of straight line is a_{1}x + b_{1}y + c_{1}z + d_{1} = 0, a_{2}x + b_{2}y + c_{2}z + d_{2} = 0 ……(1)
So, equation (1) represents straight line which is obtained by intersection of two planes.
View the video on straight lines
Equation of Straight Line in Different Forms:
Symmetrical Form:
(x–x_{1})/l = (y – y_{1})/m = (z – z_{1}) / n.
x–x_{1} / x_{2}– x_{1 }= y–y_{1} / y_{2 }– y_{1 }= z – y_{1} / z_{2} – z_{1}
x = (mx_{2} + nx_{1})/ (m+n) and y = (my_{2} + ny_{1})/ (m+n)
x/a + y/b = 1
Illustration: Find the equations of the straight lines through the point (a, b, c) which are
(a) parallel to z-axis (b) perpendicular to z-axis
Solution: (i) Equation of straight lines parallel to z-axis have
α = 90^{0}, β = 90^{0}, γ = 0^{0}
=> l = 0, m = 0, n = 1.
Therefore, the equation of straight line which is parallel to z-axis and passing through (a, b, c) is
(x – a) / 0 = (y – b) / 0 = (z – c) / 1
(ii) Equation of straight line perpendicular to z-axis
Let us assume that it makes an angle of α and β with x and y axes respectively.
Then the equation of straight line perpendicular to z axis and passing through (a, b, c) is
(x–a) / cos α = (y – b) / sin α = (z – c) / 0
=> (x–a) / l = (y – b) / m = (z – c) / 0.
Illustration: Find the coordinates of the point where the line joining the points (2, –3, 1) and (3, –4, –5) cuts the plane 2x + y + z = 7.
Solution: The direction ratios of the line are 3 – 2, –4 – (–3), –5 – 1 i.e. 1, –1, –6
Hence equation of the line joining the given points is
(x–2) / 1 = (y + 3) / –1 = (z – 1) / – 6 = r (say)
Coordinates of any point on this line are (r + 2, –r – 3, –6r + 1).
If this point lies on the given plane 2x + y + z = 7, then
2(r + 2) + (–r – 3) + (–6r + 1) = 7 => r = –1
Coordinates of the point are (–1 + 2, –(–1) – 3, –6(–1) + 1) i.e. (1, –2, 7).
Remark: If the equation of a straight line is given in general form, it can be changed into symmetrical form. The method is described in following illustration.
Illustration: Find in symmetrical form the equations of the line
3x + 2y – z – 4 = 0 = 4x + y – 2z + 3.
Solution: The equations of the line in general form are
3x + 2y – z – 4 = 0, 4x + y – 2z + 3 = 0 ……(1)
Let l, m, n be the direction cosines of the line. Since the line is common to both the planes, it is perpendicular to the normal to both the planes.
Hence 3l + 2m – n = 0, 4l + m – 2n = 0
Solving these we get,
L /–4+1 = m–4+6 = n/3–8
i.e. 1/–3 = m/2 = n/–5 = 1/√(–3)^{2} + 2^{2 }+ (–5)^{2} = 1/√38
So, the direction cosines of the line are –3/√38, 2/√38, –5/√38.
Now to find the coordinates of a point on a line: Let us find out the point where it meets the plane z = 0. Putting z = 0 in the equation given by (1), we have
3x + 2y – 4 = 0, 4x + y + 3 = 0
Solving these, we get x = –2, y = 5
So, one point of the line is (–2, 5, 0).
∴ Equation of the line in symmetrical form is
Illustration: If P = (1, 0), Q = (-1, 0) and R = (2, 0) are three given points then locus of the points satisfying the relation SQ^{2 }+ SR^{2} = 2SP^{2}, is ….? (1988)
Solution: Let the coordinate of S be (x, y).
Then, according to given condition SQ^{2 }+ SR^{2} = 2SP^{2}
Since the coordinates are given, we substitute them in the condition and obtain
(x+1)^{2} + y^{2} + (x-2)^{2 }+ y^{2} = 2[(x-1)^{2} + y^{2}]
This gives, x^{2} + 2x + 1 + y^{2} + x^{2} - 4x + 4 + y^{2} = 2 (x^{2 }+ 1 - 2x + y^{2})
This yields 2x + 3 = 0.
Hence, x = -3/2.
Hence, the line satisfying the given condition is a straight line parallel to y-axis.
Ilustration: Let a and b be non-zero real numbers then what does the equation
(ax^{2} + by^{2 }+ c)(x^{2 }- 5xy + 6y^{2}) represent? (2008)
Solution: let a and b be two non-zero real numbers.
Hence, the equation (ax^{2} + by^{2 }+ c)(x^{2 }- 5xy + 6y^{2}) implies that either
(x^{2 }- 5xy + 6y^{2}) = 0
This gives, (x-2y)(x-3y) = 0
Or x = 2y and x = 3y
These represent two straight lines passing through origin or
(ax^{2} + by^{2 }+ c) = 0 when c = 0 and a and b are of same signs then
(ax^{2} + by^{2}+c) = 0 and x= 0 and y = 0
This is actually a point specified at the origin.
When a = b and c is of opposite sign to that of a, (ax^{2} + by^{2}+c) represents a circle. Hence the given equation may represent two straight lines and a circle.
askIITians is an online portal which acts as a platform for the JEE aspirants where they can ask any kind of questions on 3D like the equation of y-axis in 3D, demonstration of the general form of straight line or equation of z-axis in 3D.
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