Foot of the Perpendicular, Distance and Image

Transcript

0:00Introduction

Hello everyone! Imagine a long, straight railway track, and you are standing in a field away from it. You want to reach the track by the shortest path. Which point of the track should you walk to? Straight across, at right angles, right? Now imagine the track is a mirror line. Where would your reflection be? Keep that question in mind. Today, in three dimensions, we find the foot of the perpendicular, the distance, and the image of a point in a line. They come up often in JEE, and one simple idea does all three. So, let's begin.

0:36The picture: point, line, foot and image

So, here is our picture. The line is vector r equals vector a plus t times vector b. The little arrows on top mean vectors, okay? Vector a is the position vector of the point A on the line, vector b, with the arrowhead, is its direction, and t is a number that slides along it. And P is a point that is not on the line. With me? Now, back to our railway track. Of all the points on this line, which one is closest to the point P? Yes, exactly, the point where the perpendicular from P meets the line! We call it N, the foot of the perpendicular. The upside down T symbol means perpendicular, and the small square in the figure marks the right angle. So P N is perpendicular to vector b, and the length P N is the distance of P from the line. Clear? Now the mirror. P prime, with the small dash, is the image of P in the line, at the end of the dashed segment. Just like in a mirror, N is exactly the midpoint of P and P prime. Right?

1:39Finding the foot: only one unknown

Now, how do we find N? N lies on the line, so N equals vector a plus t times vector b, for some value of t. This is the general point of the line, with just one unknown, t, okay? Look at the right angle in the figure again. Which condition will fix the value of t? Yes, the right angle itself! N minus P, the vector from P to N, must be perpendicular to vector b. So their dot product, the small dot on screen, is zero. Got it? Put in the general point and solve. t equals P minus vector a, dot vector b, over the modulus of vector b, squared. The vertical bars mean the modulus, the length, and the small two on top means squared. Please note the square here, because dividing by just the modulus is a common slip. Clear?

2:34Distance and image

Now the distance. It is the length of P N, and there is a direct formula for it. P N equals the modulus of P minus vector a, cross vector b, over the modulus of vector b. The small cross is the cross product. Here P minus vector a is the vector from A to P in the figure. Here is a topper's trick: if only the distance is asked, skip the foot and use this formula. It saves real time. With me? Now the image. N is the midpoint of P and P prime, so N equals one half of P plus P prime, okay? Now make P prime the subject. Many students write two P minus N for it. Is that right, or is it two N minus P? The answer is two N minus P, as you see now. Multiply the midpoint line by two, then move P across to the other side. Two P minus N would be the reflection of N in P, the wrong way round! So always double the foot, then subtract the point. Got it?

3:33Example 1: foot, image and distance from (1, 6, 3) to a line

Now, Example one from the course. Find the foot of the perpendicular from the point one, six, three, to the line x over one equals y minus one over two equals z minus two over three. Then find the image, and the distance. Pause the video and try it first. Done? Okay, let's check. Read the line first. Its point, vector a, is zero, one, two, and its direction, vector b, is one, two, three, from the denominators. So the general point N is t, then one plus two times t, then two plus three times t. Okay? Now subtract the point one, six, three, and N minus P is t minus one, then two times t minus five, then three times t minus one. Right? Now, this vector, dotted with the direction one, two, three, must give us what? Zero, yes, because they are perpendicular! So the first bracket, plus two times the second, plus three times the third, equals zero. Collect the terms, and fourteen times t, minus fourteen, equals zero. So t equals one; the double arrow means it follows. Clear? Put t equals one into the general point, and the foot N is one, three, five. That is the point of the track you would walk to! Got it? Now the image. Double the foot, then subtract the point, just as we said. Two minus one, six minus six, and ten minus three. So the image, the reflection, is one, zero, seven. Okay? And the distance is the modulus of N minus P, that is, of zero, minus three, two. It is the square root of zero plus nine plus four, so root thirteen; the tick shaped sign on screen is the square root sign. Three answers from just one value of t. Neat, right?

5:20Example 2: distance of (2, 3, 4) by the cross product

Now a harder one, Example two. Find the distance of the point two, three, four from the line x plus three over three equals y minus two over six equals z over two. Which route is faster here? Pause the video and try it first. Done? Okay, let's check. First we need a point on the line. Many students read it straight off and write three, two, zero. Is that right? No, no, careful, that is the trap! Look, x plus three means x minus minus three, so the point is minus three, two, zero. And the direction is three, six, two, from the denominators. Got it? So we use the cross product formula, no foot needed. P minus vector a is two plus three, three minus two, and four minus zero, which is five, one, four. Clear? Now the cross product of five, one, four with three, six, two; here the small dot just means times. First, one times two minus four times six. Then four times three minus five times two. Then five times six minus one times three. With me? That gives two minus twenty four, twelve minus ten, and thirty minus three. So the cross product is minus twenty two, then two, then twenty seven. Okay? Now the modulus of vector b. The square root of nine plus thirty six plus four is seven. Nice and clean, right? Finally, the distance. Squaring the three components gives four hundred and eighty four, plus four, plus seven hundred and twenty nine. Under the root, that adds to one thousand two hundred and seventeen. So the distance is root one thousand two hundred and seventeen, over seven. And we never needed the foot at all!

7:04Quick revision: foot, distance formula, image 2N minus P

So, quick revision. For the foot, write the general point, vector a plus t times vector b, and make N minus P perpendicular to vector b. For the distance, take the modulus of P minus vector a, cross vector b, over the modulus of vector b, with no foot needed. And for the image, double the foot and subtract the point, since N is the midpoint. Now go and try the two practice problems on this page. All the best!