Mixed Problems: Set 2 (Planes)

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Take two planes: three x plus four y plus twelve z equals thirteen, and six x plus eight y plus twenty four z equals thirteen. Both equations end in thirteen. So, are they the same plane, at zero distance? Keep your answer in mind. Today we solve a mixed set on planes, about two minutes per problem, and three small tools do almost all the work. Questions like these come up often in JEE Main, so let's begin.

0:36Tool 1: the normal decides the tilt

So, tool one. 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, in round brackets, read from the coefficients. The little arrow on n means a vector, and it is perpendicular to the plane. Okay? Now, let me ask you this. If two planes are parallel, what stays the same in their equations? Yes, exactly, the direction of the normal! So we can always write a parallel plane with the same a, b and c, and only the constant changes. So every parallel plane can be written as a times x, plus b times y, plus c times z, equals k. Right? Here is a quick one from the set: the plane through the point one, two, three, parallel to two x minus y plus z equals five. Keep two x minus y plus z, and put the point in to get k: two, minus two, plus three, which gives three. So the plane is two x minus y plus z equals three. Clear? And the angle between two planes is the angle between their normals; theta, the Greek letter, is that angle. Cos theta equals the modulus of, n one dot n two, over the modulus of n one, times the modulus of n two. The small ones and twos are subscripts, the dot is the dot product, and the vertical bars, the modulus, give the acute angle. Got it?

2:01Tool 2: distances

Now, tool two, distances. For the point x one, y one, z one, and the plane written equal to zero, the distance 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, and the bottom is just the length of the normal. Okay? Another quick one: the distance of the point one, one, one from two x plus two y minus z equals twelve. Move the twelve across first. Two plus two minus one minus twelve gives minus nine, and its modulus is nine. The root of four plus four plus one is three, and nine over three gives three. Right? Now two parallel planes with the same a, b, c, and constants d one and d two. The distance is the modulus of, d one minus d two, over the same square root. With me? So, can we use this formula on any two parallel planes straight away? No, wait, there is a condition here. It works only when a, b, c are exactly the same in both equations; if not, divide one equation first to make them match. Keep this in mind for our puzzle. Clear?

3:19Tool 3: planes through a line of intersection

Now tool three, with the course's picture. The two blue sheets are the planes P one equals zero and P two equals zero, each written with zero on the right. They meet in the black line, the line of intersection. Okay? Now suppose a question wants a plane through this line. Must we first find the equation of the line itself? No, and that is the beauty of it! At any point of the line, P one and P two are both zero. So P one plus lambda times P two is also zero there, for every value of lambda, the Greek letter. Each lambda gives a plane through the line, like the orange and grey sheets. Right? Change lambda, and the plane turns about the line, like a page. We get every plane through the line except P two itself, as the brackets say. With me? There is just one unknown, lambda, so one extra condition fixes it. Okay? The most common one is a point: substitute it, and solve for lambda. Got it?

4:20Example 1: distance between 3x+4y+12z=13 and 6x+8y+24z=13

Now, Example one, marked easy, is our opening puzzle. Find the distance between the planes three x plus four y plus twelve z equals thirteen, and six x plus eight y plus twenty four z equals thirteen. Pause the video and try it first. Done? Okay, let's check. Both equations end in thirteen, and thirteen minus thirteen is zero. So the planes are the same, and the distance is zero, right? Oh no, careful, that is the trap! Look at the normals first: three, four, twelve, and six, eight, twenty four. The second is two times the first, so the planes are parallel, but the coefficients are not the same. So the formula cannot be used yet. Okay? So, divide the second equation by two. It becomes three x plus four y plus twelve z equals six point five. Now both have the same a, b, c. Right? The length of the normal is the root of, nine plus sixteen plus one hundred and forty four. That is root one hundred and sixty nine, which is thirteen. Clear? With both constants on the right, the modulus of their difference is thirteen minus six point five, which gives six point five. And six point five over thirteen is one half. So the planes are half a unit apart, not the same plane, and our puzzle is solved. Got it?

5:45Example 2: plane through line of intersection and (1,1,1)

Now a harder one, Example two, marked medium. Find the plane through the line of intersection of x plus y plus z equals one and x minus y plus two z equals three, and through the point one, one, one. Pause the video and try it, with a hint: use tool three. Done? Okay, let's check. Write each plane with zero on the right: x plus y plus z minus one, and x minus y plus two z minus three. So the family is the first bracket, plus lambda times the second, equals zero. If you drop the minus one and the minus three, these planes no longer pass through our line. With me? Now, there is one unknown, lambda. Which condition fixes it here? Correct, it is the point one, one, one! In the first bracket, one plus one plus one minus one gives two. In the second, one minus one plus two minus three gives minus one. So two minus lambda equals zero, and the double arrow means which gives: lambda equals two. Clear? Now put lambda equals two back into the family: the first bracket, plus two times the second, equals zero. Okay? Collect the terms. The x terms give three x, the y terms give minus y, the z terms give five z, and the constants give minus seven. So the plane is three x minus y plus five z equals seven. Right? And now a ten second check with one, one, one. Three minus one plus five gives seven, so the point really lies on our plane. Done well within two minutes, very good!

7:27Quick revision: normal, distances, P1 + lambda P2 = 0

So, quick revision. Read the normal, a, b, c: parallel planes keep it, and the angle between planes uses the normals. For a distance, write the plane equal to zero, take the modulus, and divide by the root of a squared plus b squared plus c squared; for parallel planes, make a, b, c the same first. And through a line of intersection, write P one plus lambda times P two equals zero, and one condition fixes lambda. Now go and try the practice problem on this page. All the best!