Mixed Problems: Set 3 (Lines with Planes)

Transcript

0:00Introduction

Hello everyone! Let us start with a small picture. Put a pencil flat on your desk, so it is parallel to the desk. Now stand a pole straight up from the desk. What is the angle between the pencil and the pole? Keep your answer in mind. Today we solve a mixed set on lines with planes: angles, meeting points, and lines lying inside planes. And you will see that most of these problems come down to one dot product. These come up often in JEE Main, so let's begin.

0:33Tool 1: read the line and the plane

So, tool one: reading the line. The line is 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 its direction, vector b, is a, b, c, written in round brackets, straight from the denominators. The little arrow on top of b means a vector. Right? Now the plane. Its normal, vector n, stands straight out of the plane, and it is simply the coefficients of x, y and z. That normal is the pole from our desk picture. Clear? And where does the line meet the plane? Set each fraction equal to t, and the general point is x one plus a times t, y one plus b times t, z one plus c times t. Put it in the plane's equation and solve for t. One unknown, one equation. With me?

1:29Tool 2: the angle between a line and a plane

Now tool two, the angle. We call the angle between the line and the plane phi, the Greek letter you see on screen. Okay? Now, the dot product of vector b and vector n gives the cosine of the angle between them. But let me ask you this: is that angle the same as phi? No, it is not, because the normal stands at ninety degrees to the plane, and the small circle on screen means degrees. So the acute angle between the line and the normal is ninety degrees minus phi, the complement of phi. That is why we take the modulus of the dot product below. Right? And the cosine of ninety degrees minus phi is the sine of phi. 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 modulus: around the dot product they keep the answer positive, and around a vector they give its length. So for a line and a plane, it is sine. Got it?

2:28Tool 3: parallel, in the plane, perpendicular

Now tool three, and back to our desk picture. The pencil is parallel to the desk, and the pole is the normal. So, if a line is parallel to a plane, then its direction is parallel to the normal, right? No, no, careful, that is the trap! Remember the pencil and the pole: they make ninety degrees. That was the answer to our opening question. So a line parallel to the plane is perpendicular to the normal, and vector b dot vector n equals zero. Okay? Now, a pencil can be parallel to the desk while floating above it. So tell me, is vector b dot vector n equals zero enough for the line to lie in the plane? It is not enough, because we also need one point of the line on the plane. So the line lies in the plane when vector b dot vector n is zero, and the point x one, y one, z one satisfies the plane's equation. Both conditions, always. Clear? And the other extreme: a line perpendicular to the plane runs along the normal. So vector b is parallel to vector n, and the two small vertical lines on screen mean parallel.

3:35Example 1: sine of angle, x/2 = y/2 = z and x + 2y - 2z = 3

With me? Now Example one, the first problem of Set three, marked easy. Find the sine of the angle between the line x over two equals y over two equals z over one, and the plane x plus two y, minus two z, equals three. Pause the video and try it first. Done? Okay, let's check. x over two means x minus zero, over two, so the denominators give vector b equals two, two, one. Its modulus is the square root of four plus four plus one, which is root nine, which gives three. Okay? The normal comes from the coefficients, so vector n equals one, two, minus two. Its modulus is the square root of one plus four plus four, which again gives three. Right? Now the dot product: two times one, plus two times two, plus one times minus two. That is two plus four minus two, which gives four. Clear? So four over three times three is four over nine, and that makes cosine phi equal to four over nine, right? Oh no, careful, that fraction is the cosine of the angle with the normal, not with the plane. For the angle with the plane, the same fraction is the sine. So sine phi equals the modulus of two plus four minus two, over three times three, and the dot between the threes means times. That is four over nine. Got it?

4:57Example 2: line lies in ax + y + 2z = b, find a and b

Now a harder one, Example two, the fifth problem of Set three, marked medium. The line x minus one over two equals y minus three over one equals z minus two over minus one lies in the plane a times x, plus y, plus two z, equals b. Find the values of a and b in this plane. Pause the video and try it first. Done? Okay, let's check. First, careful with the letters: here b is just a number in the plane's equation, not the direction vector. The direction is two, one, minus one, from the denominators; the normal is a, one, two; and the line passes through one, three, two. Okay? Now, the line lies in the plane. Which two conditions does that give us? Yes, exactly, the two conditions of tool three! Condition one: the direction is perpendicular to the normal. Two times a, plus one times one, plus two times minus one, equals zero. So two times a, plus one, minus two, equals zero. The double arrow on screen means which gives, so we get a equals one half. Right? Now many students stop here. The direction is perpendicular to the normal, so the line lies in the plane for any value of b, right? No, no, wait, that only makes the line parallel to the plane, like the pencil floating above the desk. Condition two: the point one, three, two must satisfy the plane. So a times one, plus three, plus two times two, equals b. That is a, plus three, plus four, so b equals one half plus seven, which gives fifteen over two. Clear? Now a quick check. With a equals one half and b equals fifteen over two, the plane is one half x, plus y, plus two z, equals fifteen over two. Multiply by two, and that is x plus two y, plus four z, equals fifteen. With me? Take another point of the line, with t equals one, which gives three, four, one. Put it in: three plus eight plus four gives fifteen. So this point sits on the plane too, and the whole line lies in it. That's it!

7:10Quick revision: sine formula, parallel vs lying in the plane

So, quick revision. Angle between a line and a plane: use sine, with vector b and the normal, vector n. Parallel to the plane means vector b dot vector n equals zero, and lying in the plane needs a point of the line on the plane as well. And for the meeting point, put the general point into the plane and solve for t. Now go and try the practice problem on this page. All the best!