Family of Planes Through a Line

Transcript

0:00Introduction

Hello everyone! Open a book slowly. Every page turns about one line, the spine. Space works the same way: infinitely many planes pass through one straight line. So, can one single equation hold all of them? Yes, it can, with just one extra letter. That is the family of planes through a line. It comes up often in JEE, so let's begin.

0:25Every plane through the line of intersection

So, here is the picture. The first plane is P one equals zero, where P one stands for a one times x, plus b one times y, plus c one times z, plus d one. The small one below each letter means first plane. Okay? The second plane is P two equals zero, with a two, b two, c two and d two. They meet in the black line of the figure, the line of intersection. Right? Now, take any point on that line, and put it into P one and into P two. What values do you get? Both are zero, yes, exactly, because the point lies on both planes! So P one plus lambda times P two is zero for every lambda, the Greek letter, any real number. And it is still first degree in x, y and z, so it is a plane through our line, like the orange one in the figure. Clear? Change lambda, and the plane turns about the line, like a page; the grey plane in the figure is just another value of lambda. So we get every plane through the line, except P two itself, because P one never drops out of the sum. Remember that; it comes back as a trap. Got it? For a line in Cartesian form, with three equal fractions, write two planes. x minus x one over a equals y minus y one over b, and y minus y one over b equals z minus z one over c, with a, b, c the direction ratios. These are your P one and P two. With me?

1:59One extra condition fixes λ

Now, the family has just one unknown, lambda, so one extra condition usually picks out the plane. Most conditions need the normal, so collect the x, y and z terms. Vector n, with the arrow on top, has components a one plus lambda times a two, then b one plus lambda times b two, then c one plus lambda times c two. Okay? The first condition: the plane passes through a given point. Substitute it and solve for lambda. Simple, right? Second, parallel to a line with direction ratios vector b. That line is perpendicular to the normal, so vector n dot vector b is zero, and the small dot is the dot product. Clear? Third, the plane is perpendicular to a given plane. Then the two normals are perpendicular, so again their dot product is zero. With me? Fourth, parallel to an axis, say the x axis. Then the normal is perpendicular to the x axis, so the coefficient of x is zero, and the same goes for the other axes. Got it? And last, at a given distance from a point: use the distance formula, set it equal to the given distance, and solve for lambda. Careful, the modulus makes this a quadratic in lambda, so you can get two planes. Okay?

3:16Rotating a plane about the line

Now a classic use. Rotate the plane P one through a right angle, ninety degrees, with the small circle for degrees, about its line of intersection with P two. The rotated plane still contains that line. So which family must it belong to? Yes, exactly, our family, P one plus lambda times P two equals zero! And it is at right angles to P one, so its normal, vector n, is perpendicular to vector n one, the normal of P one. The upside down T is the perpendicular sign. Right? But wait, what if P one and P two are already perpendicular to each other? Then the rotated position is simply P two itself! And our family never gives P two, so no lambda works. So always check n one dot n two first, with vector n two the normal of P two; if it is zero, the answer is P two. Got it?

4:12Example 1: plane through the line and the point (1, 1, 1)

Now, Example one. Find the plane through the intersection of x plus y plus z equals six and two x plus three y plus four z plus five equals zero, and through the point one, one, one. Pause the video and try it first. Done? Okay, let's check. First bring the six across, and careful, it becomes minus six. So the family is x plus y plus z minus six, plus lambda times, two x plus three y plus four z plus five, equals zero. Right? The plane passes through one, one, one, so put x, y and z all equal to one. The brackets become one plus one plus one minus six, and two plus three plus four plus five. So what is lambda? The first bracket is minus three, and the second is fourteen. So minus three plus fourteen lambda equals zero, and the double arrow means implies: lambda is three over fourteen. Clear? Fractions are messy, so multiply the whole equation by fourteen. That gives fourteen times the first bracket, plus three times the second, equals zero. With me? Now collect the terms. Fourteen x plus six x is twenty x, and the same way we get twenty three y and twenty six z. The constants give minus eighty four plus fifteen, which is minus sixty nine, and we move it to the right side. So the answer is twenty x plus twenty three y plus twenty six z equals sixty nine.

5:45Example 2: rotate x + 2y + 3z = 4 through 90 degrees

Now a harder one, Example two. The plane x plus two y plus three z equals four is rotated through a right angle about its line of intersection with two x plus y minus z plus five equals zero. Find its new equation. Pause the video and try it first. Done? Okay, let's check. Before writing any family, what must we check first? Yes, exactly, whether the two planes are already perpendicular! Vector n one is one, two, three, and vector n two is two, one, minus one. Their dot product is two, plus two, minus three, which gives one, and the slashed equals sign says it is not equal to zero. So the family will work. Got it? Now the family, with the four moved across. x plus two y plus three z minus four, plus lambda times, two x plus y minus z plus five, equals zero. Right? Collect the coefficients of x, y and z, and the normal is one plus two lambda, then two plus lambda, then three minus lambda. Okay? Now, this normal must be perpendicular to which one, n one or n two? It must be perpendicular to n one, because we are rotating P one, so don't use n two here! We dot it with one, two, three and get one plus two lambda, plus two times, two plus lambda, plus three times, three minus lambda. That gives fourteen plus lambda equals zero, so lambda is minus fourteen. Clear? Put lambda equals minus fourteen back. The coefficients become minus twenty seven, minus twelve and seventeen, and the constant, minus four minus seventy, is minus seventy four. Multiply by minus one, and the new plane is twenty seven x plus twelve y minus seventeen z plus seventy four equals zero. With me? And a quick check against n one: twenty seven, plus twenty four, minus fifty one is zero, and the tick mark says so. The new plane really is perpendicular to the old one, beautiful!

7:58Quick revision: P1 + lambda P2, fixing lambda, rotation trap

So, quick revision. Every plane through the line of intersection is P one plus lambda times P two equals zero, all except P two itself. One extra condition fixes lambda, and most conditions use the normal. For a right angle rotation, the new normal is perpendicular to n one, and check n one dot n two first. Now go and try the two practice problems on this page. All the best!