Addition and Subtraction of Vectors
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Addition and subtraction of vectors for JEE: the triangle and parallelogram laws, the resultant formula R = root of (A squared + B squared + 2AB cos theta) and its direction, the range |A - B| to A + B, subtraction by reversing a vector, the equal-vectors and polygon shortcuts, and two solved problems: two 10 N forces at 120 degrees, and vectors of 5 and 3 at 60 degrees.
Chapters
- 0:00Introduction
- 0:30Adding two vectors: head to tail
- 1:34Magnitude and direction of the resultant R
- 2:26How big can the resultant be?
- 3:32Subtraction: reverse, then add
- 4:15Two shortcuts: equal vectors and polygons
- 5:08Example 1: resultant of two 10 N forces at 120 degrees
- 6:08Example 2: resultant of vectors 5 and 3 at 60 degrees
- 7:43Quick revision: resultant formula, subtraction, range
Transcript
0:00Introduction
Hello everyone! Okay, one small puzzle before we start. Two forces, ten newtons each, pull on the same box. What is the total force on the box? Twenty newtons, right? Not so fast! In physics, ten newtons plus ten newtons can give exactly ten newtons. Hold on to that thought. Forces, velocities and displacements all add this way, and it comes up often in JEE physics. So, let's begin.
0:30Adding two vectors: head to tail
So, look at the figure. The blue arrow along the bottom is A vector, and the orange arrow is B vector. The small arrow on top of each letter says, this is a vector, okay? How do we add them? Head to tail. Place the tail of B vector at the head of A vector, the pointed end. The dashed line on the right shows B vector moved there. This is the triangle law. Now, the total of the two vectors is called the resultant. Tell me, where does the resultant start, and where does it end? Yes, exactly! From the tail of A vector, straight to the head of B vector. That is the green arrow, R vector. R vector equals A vector plus B vector. It closes the triangle, so we call it the triangle law. Clear? Now see the same figure another way. Draw both vectors tail to tail from one corner, and complete the parallelogram with the dashed lines. Then R vector is simply the diagonal. This is the parallelogram law: same answer, different picture, okay?
1:34Magnitude and direction of the resultant R
Now, the size of R vector. The Greek letter theta, the black arc in the figure, is the angle between A vector and B vector, both drawn from the same point. Plain letters without arrows, like R, A and B here, mean just the sizes, okay? Now the formula: R equals the square root of, A squared, plus B squared, plus two A B cos theta. Write this one down. Got it? Next, the direction. The Greek letter alpha, the green arc, is the angle that R vector makes with A vector. And tan alpha equals B sin theta, divided by, A plus B cos theta. Look at the grey labels in the figure: along A vector, the resultant reaches A plus B cos theta, and across it, up the dotted line, it rises B sin theta. So tan alpha is just across, divided by along. With me?
2:26How big can the resultant be?
Now, a question for you. Theta can be anything from zero to one hundred eighty degrees. At what angle is the resultant the largest? Zero degrees, correct! The small circle means degrees, okay? Both vectors then point the same way. Cos zero is one, so R becomes the square root of, A plus B, whole squared. That is just A plus B, the largest possible. And the smallest? Opposite directions, one hundred eighty degrees. Cos is minus one, so R is the square root of, A minus B, whole squared. Oh no, careful! If you write A minus B without bars, and B is bigger, you get a negative size. So we write mod of A minus B, okay? The two straight bars mean the size, always positive. So every resultant lies in this range: mod of A minus B, is less than or equal to R, which is less than or equal to A plus B, right? The sign with a little line under it means, less than or equal to. Topper's trick: after every calculation, check that your R lies in this range. Clear?
3:32Subtraction: reverse, then add
Now, subtraction. Really simple. So, A vector minus B vector is A vector plus, in brackets, minus B vector. Minus B vector is B vector turned around: same size, opposite direction, okay? To subtract, reverse B vector, then add as before. So, tell me, what do you think changes in the formula? Only one sign! Mod of A vector minus B vector equals the square root of, A squared, plus B squared, minus two A B cos theta. Theta is still the angle between A vector and B vector, okay? Oh no, careful here! Students often keep the plus sign. Plus for adding, minus for subtracting. Got it?
4:15Two shortcuts: equal vectors and polygons
Now, two shortcuts. First, when both vectors have the same size, call that size A, okay? Then R equals two A cos of theta by two. Half the angle. And the direction? Along the bisector, the line that cuts theta into two equal halves. It makes sense: equal pulls, so the resultant sits right in the middle. Right? Second shortcut, for many vectors. Join them all, head to tail, one after another. The resultant goes from the first tail to the last head, and closes the polygon. This is the polygon law, okay? And if the last head lands back on the first tail? The polygon is already closed, so the sum is the zero vector, a zero with a small arrow on top. The double arrow on screen means, so. Vectors forming a closed polygon add to zero. With me?
5:08Example 1: resultant of two 10 N forces at 120 degrees
So, remember our puzzle? Let's finish it. Two forces of ten newtons each act at one hundred twenty degrees to each other. Find the resultant. The capital N is newton, the unit of force, okay? Pause the video, and try it first. Done? Okay, let's check. Ten plus ten, so twenty newtons, right? No, no, wait! That is the trap. Twenty only when they point the same way. Both forces are equal, so use the shortcut. R equals two times ten, cos of one hundred twenty by two, that is twenty cos sixty degrees. And cos sixty is one half. So R equals twenty times one half: ten newtons! There it is. Ten newtons plus ten newtons gives ten newtons. Puzzle solved! Direction? Equal forces, so along the bisector. Half of one hundred twenty is sixty, so the resultant is at sixty degrees to each force. Perfect!
6:08Example 2: resultant of vectors 5 and 3 at 60 degrees
Now a slightly harder one. Mod A vector is five, mod B vector is three, and the angle between them is sixty degrees. Find mod of A vector plus B vector, and its angle with A vector. Pause the video, and try it. Done? Okay, let's check. The sizes are not equal, so we use the full formula. R squared equals five squared, plus three squared, plus two times five times three, cos sixty degrees. Five squared is twenty five, three squared is nine, and two times five times three is thirty; times cos sixty, one half, that gives fifteen. Total, forty nine. So R equals seven. Very good! Now, our topper's check. Does seven pass the range check? Yes! Mod of five minus three is two, five plus three is eight, and seven sits right between them. Correct! Now the direction. Tan alpha equals three sin sixty, divided by, five plus three cos sixty. Sin sixty is root three by two, so the top is three root three by two. The bottom is five plus three by two, thirteen by two. The halves cancel: tan alpha is three root three by thirteen. Clear? So alpha is tan inverse of three root three by thirteen, the angle whose tan is this number. That is about twenty one point eight degrees; the wavy equals sign means, approximately. And this angle is measured from A vector, not from B vector. Perfect!
7:43Quick revision: resultant formula, subtraction, range
So, quick revision. To add, join head to tail, and R is the square root of, A squared, plus B squared, plus two A B cos theta. To subtract, reverse B vector and add; the cos theta term turns minus. And the resultant always lies between mod of A minus B, and A plus B; a closed polygon adds to zero. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!
More videos in Unit 2 · Kinematics
- Scalars, Vectors and Displacement
- Components and Unit Vectors
- The Dot Product
- The Cross Product
- Vectors in the Description of Motion
- Speed, Velocity and their Averages
- Uniformly Accelerated Motion
- Motion under Gravity
- Two Bodies and Two Stages
- Motion Graphs
- Variable Acceleration
- Motion in a Plane with Constant Acceleration
- Projectile Motion from the Ground
- Velocity at an Instant
- Horizontal Projection and Projection from a Height
- Projectiles on an Inclined Plane
- Relative Velocity in One Dimension
- Relative Velocity in Two Dimensions: Rain and Wind
- Crossing a River
- Closest Approach and Constrained Motion
- Angular Variables
- Centripetal Acceleration
- Non-uniform Circular Motion
- Overview
- Mixed Problems: Set 1 (Straight Line and Graphs)
- Mixed Problems: Set 2 (Vectors and Projectiles)
- Mixed Problems: Set 3 (Relative and Circular Motion)
- Overview
- Part A: JEE Main Level
- Part C: Concept Checks