Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take two lines: x equals y equals z, and x minus one equals y equals z plus one. For the shortest distance, you use the skew lines formula: the modulus of the scalar triple product, over the modulus of b one cross b two. And you get zero over zero! So is the distance zero, or is the formula broken? Hold on to your answer; we settle it in Example two. This chapter is a set of mixed two minute problems in three dimensional geometry, in the JEE Main style. So let us see how to use it, and pick up one trick on the way. Let's begin!

0:43How to use this chapter

So, first, why this chapter matters. Three dimensional geometry comes up regularly in JEE Main, usually as one multiple choice question and one numerical question. So speed and care both count. Okay? Now, the questions they often ask. The foot of a perpendicular, or an image, which means the mirror image of a point. Or the shortest distance between two lines. Right? Or a plane through a given line, or a distance measured along a given direction. So four common types in all. Clear? Now, our puzzle from the start, about two lines: which of these types is it? It is a shortest distance question, yes, exactly! Now, the chapter has three sets. Set one is points and lines, Set two is planes, and Set three is lines with planes. With me? Each set has eight solved problems, and each one should take you about two minutes. So here is a quick question for you: three sets of eight, at two minutes each. How long is the whole chapter? Three times eight is twenty four problems, and twenty four times two gives forty eight; the cross on screen means times. So about forty eight minutes in all. Keep a clock running, set by set, just like the real exam. Got it? And the real JEE Main questions, from twenty twenty one to twenty twenty six, are at the end of Chapters one to three; the short dash means to. The problems in this chapter are original, written in the same style.

2:17Topper's trick: the general point

Okay? Now, a trick that toppers use in this chapter. Here is a problem from Set three: where does the line x minus one over one, equals y minus two over two, equals z minus three over two, meet the plane two x minus y plus z equals eleven? How would you start? Set each fraction equal to t. Then x is one plus t, y is two plus two times t, and z is three plus two times t. This is the general point: every point of the line, for some value of t. Okay? Now put the general point into the plane: two times, one plus t, then minus, two plus two t, then plus, three plus two t. The numbers give two minus two plus three, which is three, and the terms in t leave two times t. So three plus two times t equals eleven, and t equals four. Right? Put t equals four back. One plus four is five, two plus eight is ten, and three plus eight is eleven. So the line meets the plane at five, ten, eleven. Clear? And the same trick works for feet of perpendiculars and for images. Write the general point, and one condition gives t, so there is one unknown and one equation. That is how these problems fit into two minutes. With me?

3:38Example 1: foot of perpendicular from origin to 2x+3y+6z=49

Now, Example one, from Set two, marked easy. Find the foot of the perpendicular from the origin to the plane two x plus three y plus six z equals forty nine. Two minutes on the clock. Pause the video and try it first. Done? Okay, let's check. The normal comes from the coefficients, so vector n is two, three, six; the little arrow on top means a vector. The perpendicular from the origin, capital O, runs along this normal. Okay? So, using our trick: what is the general point of this perpendicular line? It starts at the origin and moves along two, three, six. So the general point is two times t, three times t, and six times t. Right? Put it into the plane. Two times two t gives four times t, three times three t gives nine times t, and six times six t gives thirty six times t. Together that is forty nine times t, which must equal forty nine, so t equals one. Clear? So the foot is two, three, six, the normal's own numbers! Check it in the plane: four plus nine plus thirty six is forty nine. Got it? Now a quick question. Is the foot from the origin always just the numbers of the normal? No, be careful here, because that would be a trap. It worked here only because four plus nine plus thirty six is exactly the forty nine on the right, so t came out as one. That is special to this plane, so always solve for t, and never guess. With me?

5:09Example 2: distance between parallel lines, the 0/0 case

Now, Example two, from Set one, marked medium, and it is our opening puzzle. Find the distance between the lines x equals y equals z, and x minus one equals y equals z plus one. Pause the video and try it first. Done? Okay, let's check. In x equals y equals z, each part is over one, so the direction, vector b, is one, one, one. And this line passes through the origin, zero, zero, zero. Okay? Now the second line, x minus one equals y equals z plus one. Which point does it pass through? It passes through one, zero, minus one, because z plus one means z minus minus one, so watch that sign. And its direction is again one, one, one. Right? Same direction, so the lines are parallel. Then b one cross b two is the zero with an arrow, the zero vector; here the cross sign means the cross product. The small numbers below b name the two lines. So the skew formula gives zero over zero, exactly our puzzle. Clear? So the distance is zero, right? No, no, wait, that is the trap! Zero over zero is not zero; it only tells us the skew formula does not apply. For parallel lines, the course uses a different formula: d equals the modulus of, vector a two minus vector a one, cross vector b, over the modulus of vector b; the vertical bars mean the modulus. Here vector a one and vector a two are the position vectors of points on the two lines. With me? Vector a one is the origin, so vector a two minus vector a one is just one, zero, minus one. Cross it with one, one, one: zero plus one, then minus one minus one, then one minus zero, which gives one, minus two, one. Its modulus is the root of one plus four plus one, which is root six; the tick shaped sign is the square root. Got it? The modulus of vector b is root three. So d is root six over root three, which is root two. Puzzle solved: not zero, and not zero over zero, but root two. So always check first whether the lines are parallel. Okay?

7:29Quick revision: timing, general point, parallel-line formula

So, quick revision. About two minutes a problem, so time yourself, set by set. For meeting points, feet and images, write the general point, use one condition, and solve for t. And when b one cross b two is the zero vector, the lines are parallel, so use the parallel line formula. Now open Set one, start your clock, and try the practice problem at the end of each set. All the best!