Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! One quick puzzle before we start. How far is the point two, three, four from the plane three x minus six y plus two z plus eleven equals zero? Three students try it: the first gets four sevenths, the second gets one seventh, and the third gets exactly one. Which one is right? Keep your answer ready; by the end, you will see where the other two slipped. This is Part A of our three dimensional geometry problem bank, at JEE Main level, and questions like this come up often. So, let's begin.

0:35Part A: a 30-minute test

So, here is what Part A holds. Eight solved problems, all at JEE Main level. Short, quick questions, okay? And they are mixed: distances, the angle between two planes, an image, a foot, a line meeting a plane, and a volume. Just like the real exam, nobody tells you the chapter. Right? The plan, from the chapter overview: after Chapters one to four, attempt Part A as a thirty minute test. First, we sharpen two tools and solve two problems together. With me?

1:10Tool 1: distance of a point from a plane

Tool number one, from Chapter three: the distance of a point from a plane. For the plane a times x, plus b times y, plus c times z, plus d, equals zero, the normal, perpendicular to the plane, is the vector n equals a, b, c. The little arrow on n means a vector. Okay? Now look at the figure. P is the point x one, y one, z one; the small ones below are subscripts. The red arrow is the normal n, and the label n dot r equals d is the plane in vector form; careful, that d is just the constant on the right there, not the d of our equation. So, from P, which path to the plane is the shortest? Yes, exactly, straight along the normal! It lands at N, the foot, where the small square marks the right angle. 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 vertical bars mean the modulus, the tick-shaped sign is the square root, and the small two means squared. P prime, with the small dash, is the mirror image. Clear? Now, a warning: the formula needs the plane written as something equals zero, and the three dots stand for the left side. If a constant sits on the right, move it across first. Got it?

2:37Tool 2: shortest distance between skew lines

Tool number two, from Chapter two. In space, two lines can be neither parallel nor intersecting; we call them skew lines. The first is vector r equals vector a one, plus lambda times vector b one. The arrows on top mean vectors, the small one and two are labels, and lambda, a Greek letter, slides along the line. The second has vector a two, vector b two, and mu, another Greek letter. In the figure, they are L one and L two. Okay? Now, join any point of L one to any point of L two. Of all such segments, the shortest one is the segment P Q, drawn in the figure. So look at it carefully and tell me, what is special about P Q? Yes, exactly, it is perpendicular to both lines! The small squares at P and at Q mark the right angles. The vector perpendicular to both b one and b two is b one cross b two; that is the small cross sign. So P Q is along it; in the figure, the two vertical lines mean parallel. Got it? Its length is the shortest distance d, also in the figure. d equals the modulus of, vector a two minus vector a one, dot, b one cross b two, over the modulus of b one cross b two. The dot is the dot product, and the top is called the scalar triple product. Clear? Now the conditions, and they really matter. If the lines are not parallel and this numerator is zero, they meet, so the distance is zero. Right? And for parallel lines, b one cross b two is the zero vector, the zero with an arrow, so this formula fails; parallel lines use another formula. So first, check whether the directions are proportional. With me?

4:23Example 1: distance of (2, 3, 4) from 3x - 6y + 2z + 11 = 0

Now, remember our puzzle? Let's settle it. Find the distance of the point two, three, four from the plane three x minus six y plus two z plus eleven equals zero. Pause the video, and try it first. Done? Okay, let's check. The plane is already written equal to zero, which is good. So a is three, b is minus six, c is two, and d is eleven. Right? Now put the point in: three times two, minus six times three, plus two times four. That gives six minus eighteen plus eight, which is minus four. So the top is four, isn't it? No, no, wait, we forgot the plus eleven! That was our first student's slip, and it gave four sevenths. Six minus eighteen plus eight plus eleven, equals seven. Okay? Now the bottom. Nine plus thirty six plus four is forty nine. So we divide seven by forty nine and get one seventh, right? Oh no, careful, the formula has a square root at the bottom! That was our second student's slip. The square root of forty nine is seven. Clear? So the distance is the modulus of seven, over seven, which is exactly one. Our third student was right, and the puzzle is solved. Perfect work if you got it!

5:48Example 2: skew lines x = y = z and x = 1, y = -z

Now the harder one. Find the shortest distance between the line x over one equals y over one equals z over one, and the line x equals one, y over one equals z over minus one. Look carefully at that x equals one, and pause the video to try it first. Done? Okay, let's check. The first line passes through the origin, so vector a one is zero, zero, zero, and its direction, vector b one, is one, one, one. Okay? Now the second line. Many students write its direction as one, one, minus one. Is that right? No, and that is exactly the trap here! x equals one means x never changes along this line, so the direction ratio along x is zero. So vector b two is zero, one, minus one, and the point one, zero, zero lies on the line. These directions are not proportional, so the lines are not parallel, and our formula applies. Got it? Now b one cross b two. The components are minus one minus one, then zero plus one, then one minus zero. So we get minus two, one, one. Right? Next, vector a two minus vector a one is one, zero, zero, and its dot product with minus two, one, one is minus two. The crossed equals sign means not equal: the numerator is not zero, so the lines do not meet. Clear? The modulus of the cross product is the square root of four plus one plus one, which is root six. With me? So the distance is minus two over root six, isn't it? No, no, wait, a distance can never be negative; that is why the formula has a modulus on top. The modulus of minus two is two, so d equals two over root six, which is the square root of two thirds. Beautiful, isn't it?

7:46Quick revision: point to plane distance and skew lines

So, quick revision. Point to plane: 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. Skew lines, not parallel: the modulus of the scalar triple product, over the modulus of b one cross b two. And read each line carefully: x equals one means the direction ratio along x is zero. Now attempt Part A as a thirty minute test, then try the practice problem on this page. All the best!