Projections, Components and Work

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. You pull a heavy trolley along a flat road, with a rope tilted upwards. You pull hard, but not all of that pull moves the trolley forward, right? Only the part along the road does the real job. And we find that part with one small idea, the projection: the shadow of one vector on another. It gives components, distances and work done, and it comes up often in JEE questions. So, let's begin.

0:30Scalar projection: the shadow of vector a on vector b

Look at the figure. From the point O we draw two vectors, a and b, each with a small arrow on top. Theta, the Greek letter, is the angle between them. Now drop a perpendicular from the tip of a vector onto the line of b. It lands at N, and the small square there means a right angle, okay? The orange arrow from O to N is the shadow of a vector on b, its projection. With me? Now, the length of this shadow. In the right triangle, the hypotenuse is mod a, the length of a vector, and the two vertical bars mean modulus. So O N equals mod a, times cos theta, as the label under the curly brace says. Right? But a JEE question usually gives components, not the angle. So we use the dot product, the small dot: a dot b equals mod a, times mod b, times cos theta. Okay? Divide both sides by the modulus of b. So O N, the scalar projection, is a dot b, over the modulus of b. And b vector over its modulus is b hat, the unit vector along b, with the little cap on top. So the projection is simply a dot b hat. The crossed equals sign means not equal: b must not be the zero vector, the zero with an arrow, since we divide by its modulus. Clear? Now let me ask you this. What happens to the scalar projection when theta is obtuse? Yes, exactly, cos theta is negative, so the scalar projection is negative. So never drop that minus sign from the scalar projection; only the length of the shadow is its absolute value. Got it?

2:14Vector projection and the perpendicular part

Now, the scalar projection is just a number. But the orange arrow is a vector along b. So how do we turn that number into the arrow? Yes, exactly, multiply it by b hat! So the vector projection is a dot b hat, in brackets, times b hat. Write each b hat as b over its modulus, and you get a dot b, over the modulus of b squared, times b vector. The small two on top means squared. Okay? Now be careful here, because this square is a classic trap. Many students write just the modulus of b in the denominator. But one modulus comes from the scalar projection, and one more from b hat, so the modulus of b is squared. Right? Take this part away from a vector, and what is left is the green arrow, from N up to the tip: the perpendicular part, a vector minus the vector projection. Every vector splits into a part along b, and a part perpendicular to b, as the right angle at N shows. With me? This green side gives a shortcut too. For the distance d of a point P from the line through a point A along b, let the point A sit at O, and P at the tip. The long arrow over A P, the vector from A to P, is the hypotenuse, and its shadow is A P dot b hat. So by Pythagoras, d is the square root of, the modulus of A P squared, minus A P dot b hat in brackets, squared. Clear?

3:45Work done by a constant force

Now, back to our trolley. When a constant force F moves a particle through a displacement d, the work done is W equals F dot d. Just one dot product, okay? Look, F dot d, over the modulus of d, is the projection of F on d, the part of the pull along the road. Multiply by the modulus of d, the size of the displacement, to get the work. So the upward part of the pull, at right angles to the road, adds nothing to the work, and our puzzle is solved, right? Now let me ask you this. The particle moves from a point A to a point B, so what is the displacement vector d? Yes, it is the vector A B, with the long arrow on top: the position vector of B, minus the position vector of A, final minus initial, always. Okay? Now, if the force is perpendicular to the displacement, how much work does it do? It does zero work, because the dot product of two perpendicular non zero vectors is zero. The upside down T means perpendicular, and the double arrow means implies. Got it?

4:54Example 1: projections of 5i + j + k on 2i - j + 2k

Now, example one. Find the scalar and vector projections of a vector, five i hat plus j hat plus k hat, on b vector, two i hat minus j hat plus two k hat. Here i hat, j hat and k hat are the unit vectors along the axes. Pause the video and try it first. Done? Okay, let's check. For a dot b, multiply matching components and add: five times two is ten, one times minus one is minus one, and one times two is two. Ten minus one plus two gives eleven. Right? Next, the modulus of b: the square root of, four plus one plus four, so three. With me? So, what is the scalar projection? That is correct, it is a dot b over the modulus of b, which is eleven over three. If you got this one, you are doing really well. Clear? Now the vector projection. Is it just eleven over three, times b vector? No, no, careful, that is the trap we just saw! We divide by the modulus of b squared, which is nine. So the vector projection is eleven over nine, times the bracket, two i hat minus j hat plus two k hat, and that's it!

6:12Example 2: work done by two forces moving a particle

Now a slightly harder one, in JEE style. Two forces act on a particle: four i hat plus j hat minus three k hat, and three i hat plus j hat minus k hat. They move it from the point i hat plus two j hat plus three k hat, to the point five i hat plus four j hat plus k hat. Find the total work done. Pause the video and try it first. Done? Alright, let's solve it. First, add the two forces to get the resultant force. Four plus three is seven, one plus one is two, and minus three minus one is minus four. So F is seven i hat plus two j hat minus four k hat. Okay? Now, is the displacement d simply the final point, five i hat plus four j hat plus k hat? No, no, careful, that is only the final position vector! Displacement is final minus initial: five minus one is four, four minus two is two, and one minus three is minus two. So d is four i hat plus two j hat minus two k hat. Right? Now the work, F dot d. Seven times four is twenty eight, two times two is four, and careful with the sign: minus four times minus two is plus eight. Twenty eight plus four plus eight, equals forty. So the total work done is forty units. See how fast that was?

7:43Quick revision: projections, perpendicular part, work

So, quick revision. The scalar projection is a dot b hat, or a dot b over the modulus of b, negative when theta is obtuse. The vector projection is a dot b over the modulus of b squared, times b vector, and the perpendicular part is a vector minus it. And work done is F dot d, where d is final minus initial position. Now go and try the practice problems on this page. All the best!