Equations of a Line
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Equation of a line in 3D for JEE: the vector form r = a + lambda b, the Cartesian symmetric form, the two-point form, what a zero denominator means, the general point that starts almost every problem, a line as the meeting of two planes, the coefficient trap, and two solved course problems.
Chapters
- 0:00Introduction
- 0:33Vector form: a point and a direction
- 1:24Cartesian form: the symmetric form
- 2:29The general point: where most problems start
- 2:58A line as the meeting of two planes
- 3:56Trap: make the coefficients of x,y,z equal to +1
- 4:56Example 1: vector and Cartesian equations through (1, 2, -3)
- 6:03Example 2: points on a line at distance 3 from the origin
- 7:40Quick revision: both forms, coefficient trap, general point
Transcript
0:00Introduction
Hello everyone! Let us start with a small picture. An ant sits at a point in space and walks along a straight path, forwards or backwards, never turning. How would you describe every point on that path? Sounds hard, right? But you need only two things: one point, and one direction. Today we turn those two things into the equations of a line in space, a topic that comes up often in JEE three dimensional geometry. So, let's begin.
0:33Vector form: a point and a direction
So, here is the vector form of a line: vector r equals vector a, plus lambda times vector b. The little arrows on top mean vectors, and lambda, the Greek letter, is just a real number. Okay? Now, vector a is the position vector of one fixed point on the line, where our ant starts. Right? And vector b is any vector parallel to the line, the direction in which the ant walks. So tell me, vector r is the position of the ant. What happens to it as lambda changes? Yes, exactly! When lambda is zero, the ant is at the start. Positive lambda moves it forwards along vector b, and negative lambda moves it backwards. Each value of lambda gives one point, and together they make the whole line. Clear?
1:24Cartesian form: the symmetric form
Now the same line without vectors: the Cartesian, or symmetric, form. x minus x one, over a, equals y minus y one, over b, equals z minus z one, over c. The small ones below the letters are subscripts, okay? It passes through the point x one, y one, z one, and a, b, c are its direction ratios, written d r's on screen. Any non-zero multiple of them gives the same direction. And unlike direction cosines, a squared plus b squared plus c squared need not be one; the small two up top means squared. Right? Now, what if one of the denominators is zero? We cannot divide by zero, so what could it mean? Look at z minus three, over zero. It means that coordinate never changes: z equals three all along the line. With me? And through two points, the direction ratios are just the differences of their coordinates: x minus x one, over x two minus x one, and the same for y and z. This is the two point form. Got it?
2:29The general point: where most problems start
Now the most useful idea of this chapter. Set each fraction equal to the same number, t. All three are equal anyway, so we just give their common value a name. Okay? Then x equals x one plus a times t, y equals y one plus b times t, and z equals z one plus c times t. Together, this is the general point of the line, and every value of t gives one point. Almost every problem on lines starts right here. Remember it!
2:58A line as the meeting of two planes
Sometimes a line is given as the meeting of two planes: vector n one dot vector r equals d one, and vector n two dot vector r equals d two. The dot is the dot product, and n one, n two are the normals, the vectors perpendicular to the planes. Okay? Now think. The line lies in both planes, so it is perpendicular to both normals. Which vector is perpendicular to two given vectors at once? Yes, exactly, it is the cross product of the two normals! So the direction of the line is n one cross n two; the cross sign means the cross product. Right? For one point on it, put one coordinate equal to zero, say z equals zero, and solve the two plane equations for x and y. If that gives no solution, the line never meets that plane, so set a different coordinate to zero instead. Point plus direction, and the line is ready!
3:56Trap: make the coefficients of x,y,z equal to +1
Now, a very common trap. Two x minus one, over three, equals y plus two, over two, equals one minus z, over four. So the direction ratios are three, two and four, right? Just read the denominators? Oh no, careful, that answer is wrong! First, the coefficients of x, y and z must be plus one. Divide the top and bottom of the first fraction by two: x minus one half, over three over two, written with a slash. And one minus z is the negative of, z minus one, so the minus sign moves down: z minus one, over minus four. Okay? Now the denominators are three over two, two and minus four. Multiply all of them by two, and the direction ratios are three, four and minus eight. Very different, right? And the point, read from the new numerators, is one half, minus two, and one. So make the coefficients plus one first, and only then read. Clear?
4:56Example 1: vector and Cartesian equations through (1, 2, -3)
Okay, let's solve some. Write the vector and Cartesian equations of the line through the point one, two, minus three, parallel to two i hat, minus j hat, plus four k hat. The little caps are hats; they mark the unit vectors along the three axes. Pause the video and try it first. Done? Okay, let's check. The point gives vector a, i hat plus two j hat minus three k hat, and vector b is the direction, two i hat minus j hat plus four k hat. Put them into the vector form. Vector r equals, i hat plus two j hat minus three k hat, plus lambda times, two i hat minus j hat plus four k hat. Easy, isn't it? Now the Cartesian form. The point is one, two, minus three, so the last fraction is z minus three, over four, right? No, be careful here, because z minus, minus three, is z plus three. So x minus one, over two, equals y minus two, over minus one, equals z plus three, over four. And keep that minus one under y. Got it?
6:03Example 2: points on a line at distance 3 from the origin
Now a harder one, and our ant is back. Find the points on the line, x plus one, over two, equals y minus one, over one, equals z over two, at a distance of three from the origin. Hint: start from the general point. Pause the video and try it first. Done? Okay, let's check. Set each fraction equal to t: x equals two t, minus one, y equals t plus one, and z equals two t. That is the general point. Okay? The distance is three, so its square is nine. Two t minus one, the whole squared, plus t plus one, the whole squared, plus four t squared, equals nine; the small two up top means squared. With me? Expand and collect: nine t squared, minus two t, minus seven, equals zero. Now solve it. The double arrow means implies, and it gives t equals one, or t equals minus seven over nine. So we have two values of t. How many points does that give us? Two points, yes, so don't stop after the first one! With t equals one, the point is one, two, two. Let's check it: one plus four plus four is nine, and its square root is three. Perfect! With t equals minus seven over nine, x is minus fourteen over nine, minus one, which gives minus twenty three over nine. y is two over nine, and z is minus fourteen over nine. So the ant is three units from the origin at exactly two places. Clear?
7:40Quick revision: both forms, coefficient trap, general point
So, quick revision. Vector form: vector r equals vector a plus lambda times vector b, and the Cartesian form has three equal fractions. Make the coefficients of x, y and z plus one before reading direction ratios. And start almost every problem from the general point. Now go and try the practice problems on this page. You can do them. All the best!
More videos in Unit 11 · Three-Dimensional Geometry
- Coordinates, Distance and Section Formulae
- Direction Cosines and Direction Ratios
- Vector Tools: Area, Volume and Coplanarity
- Angle, Intersection and Coplanarity of Two Lines
- Foot of the Perpendicular, Distance and Image
- Shortest Distance Between Two Lines
- Equations of a Plane
- Angles, Distances, Feet and Images
- Family of Planes Through a Line
- A Line and a Plane
- Overview
- Mixed Problems: Set 1 (Points and Lines)
- Mixed Problems: Set 2 (Planes)
- Mixed Problems: Set 3 (Lines with Planes)
- Overview
- Part A: JEE Main Level
- Part C: Concept Checks