A Line and a Plane

Transcript

0:00Introduction

Hello everyone! Picture a flagpole leaning on flat ground. What is the angle between the pole and the ground? Easy, just measure it, right? But the ground is a plane, and a plane has no single direction to measure from. So how do we find that angle, and why does its formula use sine, not cosine? That is our topic today, a line and a plane: the angle between them, the special cases, and where they meet. This comes up often in JEE, so let's begin.

0:31Angle between a line and a plane

So, the set up. The line is vector r equals vector a, plus t times vector b; the little arrow on top means a vector. Vector a gives a point on the line, vector b gives its direction, and the number t slides us along it. The plane is vector n dot vector r equals d, where the dot is the dot product and vector n is the normal, perpendicular to the plane. Okay? Now, the line has one direction, vector b, but the plane spreads out in every direction. So which one direction belonging to the plane should we compare the line with? Yes, exactly, it is the normal! Phi, this Greek letter, is the angle between the line and the plane, and the small circle means degrees. The line and the normal make ninety degrees minus phi, just like our flagpole: its angle with the ground, plus its angle with a vertical post, makes ninety degrees. With me? Now, the dot product of vector b and vector n, divided by their lengths, gives the cosine of the angle between vector b and the normal; take its modulus, and you get the cosine of the acute angle, ninety degrees minus phi. So what is the cosine of ninety degrees minus phi? It is sine phi, exactly! So sine phi equals the modulus of, vector b dot vector n, over the modulus of vector b, times the modulus of vector n; the vertical bars mean the modulus. The modulus keeps phi between zero and ninety degrees. And here is the JEE trap: out of habit from two lines, students write cos phi. For a line and a plane, it is sine. Got it?

2:14Parallel, lying in the plane, perpendicular

Now, the special cases. When the line is parallel to the plane, it is perpendicular to the normal, so vector b dot vector n equals zero, okay? So tell me, if vector b dot vector n is zero, does the line lie in the plane? No, not necessarily! It could float above the plane, like a pencil held flat over a table. To lie in the plane, vector a, a point of the line, must also satisfy the plane's equation. We need both conditions together. Clear? And the line is perpendicular to the plane when vector b is parallel to vector n; the two small vertical lines mean parallel. Think of a vertical post standing on flat ground. Right?

2:58Where a line meets a plane

Now, where does the line meet the plane? Every point of the line is vector a, plus t times vector b, for some value of t. We call it the general point, okay? The meeting point lies on the line and on the plane. So how do we find its value of t? Yes, put the general point into the plane's equation! Solve for t, put it back, and you have the meeting point. Simple, right? The same idea measures a distance along a given direction, vector u. From a point P, walk along vector u, through the points P plus t times vector u, until you reach the plane, okay? Now, is the distance just t? No, careful, not in general! The number t only counts how many copies of vector u we walked, and it is negative if we walk backwards. So the distance is the modulus of t, times the modulus of vector u. With me?

3:52Projection and image of a line

One last idea: the projection of a line on a plane. Let X be the point where the line meets the plane. From any other point of the line, drop a perpendicular to the plane, and join X to its foot. That is the projection, okay? For the image of the line, join X to the mirror image of another point instead. Right? But if the line is parallel to the plane, there is no meeting point. So what do we do? Simple, take the images of any two points of the line, and join them. Clear?

4:25Example 1: line meets x - y + z = 5, distance from a point

Now, Example one. Find where the line x minus two over three, equals y plus one over four, equals z minus two over two, meets the plane x minus y plus z equals five, and the distance of that point from minus one, minus five, minus ten. Pause the video and try it first. Done? Okay, let's check. Set each fraction equal to t. So the general point is two plus three t, minus one plus four t, two plus two t. Okay? Put it into the left side of the plane, and careful with the minus in front of the second bracket. The numbers give two plus one plus two, which is five, and the t terms give just t. So we get five plus t. Right? Five plus t must equal five, so t equals zero, and the meeting point is two, minus one, two. Got it? Now, how far is minus one, minus five, minus ten from two, minus one, two? The differences of the coordinates are three, four and twelve. So the distance is the square root of, three squared plus four squared plus twelve squared; the tick shaped sign is the square root, and the small two up top means squared. That is root one hundred and sixty nine, which is thirteen. Very good!

5:47Example 2: angle between a line and 10x + 2y - 11z = 3

Now a harder one, Example two. Find the angle between the line x plus one over two, equals y over three, equals z minus three over six, and the plane ten x plus two y minus eleven z equals three. Pause the video and try it first. Done? Okay, let's check. The direction comes from the denominators, so vector b is two, three, six. Its modulus is the root of four plus nine plus thirty six, that is root forty nine, which is seven. Okay? The normal comes from the coefficients, so vector n is ten, two, minus eleven. Its modulus is the root of one hundred plus four plus one hundred and twenty one, that is root two hundred and twenty five, which is fifteen. Right? Now the dot product: twenty plus six minus sixty six, which gives minus forty. Clear? Now, the moment of truth. Do we use cosine or sine here? And what happens to that minus sign? We use sine, because the angle with a plane is the complement of the angle with its normal, and the modulus turns minus forty into forty. The dot between seven and fifteen means times, giving one hundred and five. So sine phi is forty over one hundred and five, which simplifies to eight over twenty one. With me? So phi equals sine inverse of eight over twenty one; the small minus one up top means inverse, not one over sine. If you wrote cos inverse, you found the angle with the normal, not with the plane. That is exactly the trap. Got it?

7:31Quick revision: sine formula, parallel case, meeting point

So, quick revision. For the angle between a line and a plane, use sine phi, with the direction vector b and the normal vector n. Parallel means vector b dot vector n is zero; to lie in the plane, a point of the line must lie in it too. For the meeting point, put the general point into the plane and solve for t. Now try the two practice problems on this page. All the best!