Angles, Distances, Feet and Images

Transcript

0:00Introduction

Hello everyone! Stand in front of a mirror. How far behind the glass does your reflection seem to be? Exactly as far behind as you are in front, right? Now think of a plane in space as a huge flat mirror. Today we find how far a point is from a plane, where the perpendicular from it lands, and where its mirror image sits, plus the angle between two planes. These come up often in JEE, and a few short formulas do it all. So, let's begin.

0:30Angle between two planes

First, a quick reminder. The plane a times x, plus b times y, plus c times z, plus d, equals zero, has the normal vector n equals a, b, c, read straight from the coefficients. The little arrow on n means a vector, and the normal is perpendicular to the plane. Okay? Now, two planes cut each other, like two pages of a half open book. Let me ask you this: which one direction decides how a plane is tilted? Yes, exactly, the normal! Tilt the plane, and its normal tilts with it. So the angle between two planes is just the angle between their normals. Clear? Now theta, the Greek letter, is that angle. Cos theta equals the modulus of, vector n one dot vector n two, over the modulus of n one, times the modulus of n two. The small one and two below are labels, the dot is the dot product, and the vertical bars mean the modulus. We put a modulus on top because two crossing planes make an acute and an obtuse angle, and the modulus gives us the acute one. Got it? And two special cases. Parallel planes have proportional normals. For perpendicular planes, n one dot n two equals zero. With me?

1:47Distance of a point from a plane

Now, distances. Look at the figure. Here is a point P, at x one, y one, z one; the small ones below are subscripts. And our plane is a times x, plus b times y, plus c times z, plus d, equals zero. In the figure, the sheet is the plane, labelled in vector form as vector n dot vector r equals d; careful, that d is a different constant from the d in our equation. The red arrow n is its normal. The point below, with a small dash, is P prime; more on it soon. From P, which path to the plane is the shortest? Yes, exactly, straight along the normal! It lands at N, the foot of the perpendicular. And the length P N is the modulus of, a times x one, plus b times y one, plus c times z one, plus d, over the square root of, a squared plus b squared plus c squared; the tick shaped sign is the square root. So, put the point in, take the modulus, and divide by the length of the normal. Right? The same expression gives one more test. Put two points into it. The same sign means the same side of the plane, and opposite signs mean opposite sides.

3:04Two parallel planes

Okay? Now, two parallel planes, written with the same a, b, c. Only the constants change: d one for the first, and d two for the second. Right? The distance is the modulus of, d one minus d two, over the square root of, a squared plus b squared plus c squared. So, can we use it straight away on any two parallel planes? No, wait, there is a condition. It works only when a, b, c are exactly the same in both equations. If they are not the same, divide one equation first to make them match, then subtract the constants. Got it?

3:41Foot and image of a point

Now the most useful part: the foot and the image. From P, in which direction do we walk to reach the foot N? Along the normal, yes! So the line through P is: x minus x one over a, equals y minus y one over b, equals z minus z one over c, and we call this common value t. Okay? The foot N also lies on the plane, so put the general point of this line into the plane's equation and solve for t. For the foot, t equals minus, a times x one, plus b times y one, plus c times z one, plus d, over a squared plus b squared plus c squared. Look at the minus sign in front. Drop it, and you walk away from the plane! Clear? Now the mirror. The image is on the other side of the plane, like your reflection. So how far along the normal do we walk to reach it? Twice as far, exactly! That is P prime, the image. It is as far behind the plane as P is in front, just like our mirror. So double this value of t, and the foot N becomes the midpoint of P and P prime. With me? And here is a topper's trick. Put the foot back into the plane's equation. It must give zero; if it does not, there is a sign mistake somewhere. Okay?

5:03Example 1: distance of (2, 5, -3) from 6x - 3y + 2z = 4

Now, Example one. Find the distance of the point two, five, minus three from the plane six x minus three y plus two z equals four. Pause the video and try it first. Done? Okay, let's check. Many students put the point straight into six x minus three y plus two z. Twelve minus fifteen minus six gives minus nine, so the distance is nine over seven, right? No, careful, that is the trap! The formula needs the plane equal to zero, so first move the four across: six x minus three y plus two z minus four equals zero. Okay? Now put in the point: twelve, minus fifteen, minus six, minus four, which gives minus thirteen. Right? The length of the normal is the square root of thirty six plus nine plus four, which is root forty nine, and that is seven. Clear? So the distance is the modulus of minus thirteen, over seven, which is thirteen over seven. That forgotten four changes the answer, so always write the plane equal to zero first. Got it?

6:15Example 2: foot and image of (1, 3, 4) in 2x - y + z + 3 = 0

Now a harder one, Example two. Find the foot of the perpendicular and the image of the point one, three, four in the plane two x minus y plus z plus three equals zero. In our figure, P is now one, three, four. Pause the video and try it first. Done? Okay, let's check. Put P into the left side: two minus three plus four plus three, which gives six. And a squared plus b squared plus c squared is four plus one plus one, also six. With me? Now, what is t for the foot? Remember that minus sign. Yes, t equals minus six over six, which is minus one. It is negative because we walk against the normal, towards the plane. Okay? So the foot is one, three, four, minus one times the normal two, minus one, one; the dot here means times. One minus two, three plus one, four minus one. So the foot N is minus one, four, three. Got it? For the image, double t, so take away two times the normal. One minus four, three plus two, four minus two. So the image P prime is minus three, five, two. Right? Now the topper's check. Put the foot into the plane: two times minus one, minus four, plus three, plus three. That is zero, so the foot really lies on the plane, and our answer is confirmed. Very good work!

7:46Quick revision: angle, distance, foot and image formulas

So, quick revision. The angle between two planes is the angle between their normals, with a modulus for the acute angle. For distance, write the plane equal to zero, put the point in, take the modulus, and divide by the root of a squared plus b squared plus c squared. For the foot, walk along the normal by t; for the image, walk two times t. Now go and try the three practice problems on this page. All the best!