Mixed Problems: Set 1 (Points and Lines)

Transcript

0:00Introduction

Hello everyone! Let us start with a quick challenge. In a JEE Main paper, you get a point and a line in space, and the question asks for the mirror image of the point in that line. You have about two minutes. So, what is the very first thing you write? Keep your answer in mind. Today we solve a mixed set on points and lines, and you will see that one habit, one unknown and one condition, cracks these problems fast. Questions like these come up often, so let's begin.

0:33Tool 1: dividing a join in a ratio

So, tool one. Take two points in space: P, with coordinates x one, y one, z one, and Q, with x two, y two, z two. The small ones and twos below the letters are subscripts. Okay? Now the point that divides P Q internally in the ratio m to n; the colon on screen means ratio. Its x coordinate is m times x two, plus n times x one, over m plus n, and the same pattern gives y and z. Notice, m goes with x two, the far end, not with x one. Swapping them is a very common slip. Right? For external division, simply replace n by minus n. Clear? Now let us bring in a coordinate plane. Let me ask you this: on the x y plane, which coordinate is zero? Yes, exactly, z is zero! So the x y plane is z equals zero, and it divides P Q in the ratio minus z one to z two, as on screen. Just put the z coordinate of the section formula equal to zero. With me? And how do you read the sign? If the ratio comes out positive, the division is internal; if it comes out negative, it is external, because of that minus n. Got it?

1:49Tool 2: the foot and the image in a line

Now tool two, with the course's picture. The line is vector r equals vector a plus t times vector b; the little arrows on top mean vectors. Vector a takes us to the point A on the line, vector b, with the arrowhead, is the direction, and t is any real number. And P is a point off the line. Okay? Now back to our challenge. You have the point P and this line, and you want the image. What is the very first thing you write? Yes, you write the general point of the line! N equals vector a plus t times vector b. Every point of the line has this form, including N, the foot in the figure, so there is just one unknown, t. The course says almost every problem starts here. Right? Now, one unknown needs one condition. The small square in the figure marks a right angle, and the upside down T in orange means perpendicular. So N minus P, the vector from P to N, is perpendicular to vector b, and their dot product, the small dot on screen, is zero. That fixes t. Clear? And the image. P prime, with the small dash, sits at the end of the dashed segment, like a reflection in a mirror. N is the midpoint of P and P prime, so P prime equals two N minus P, as on screen. With me?

3:13Example 1: ratio in which the xy-plane divides a join

Now Example one, problem two of Set one, marked easy. In what ratio does the x y plane divide the join of the points two, four, minus three, and minus one, two, six? And then, find the point. Pause the video and try it first. Done? Okay, let's check. Let the ratio be k to one, so k is our one unknown. P is two, four, minus three, and Q is minus one, two, six. So the z coordinate of the dividing point is k times six, plus one times minus three, over k plus one. That is six times k, minus three, over k plus one. Okay? Now, one unknown needs one condition. Which condition fixes k here? Yes, exactly, the point lies on the x y plane, so its z coordinate is zero! Six times k, minus three, equals zero, and the double arrow means which gives: k equals one half. Right? So the ratio is one half to one, which is one to two. It is positive, so the division is internal. Clear? Now the quick check from tool one, minus z one to z two. Here z one is minus three, so the ratio is minus three to six, right? No, no, careful with the sign here, because z one is already minus three. So minus z one is plus three, and minus three to six would wrongly say external. Three to six is one to two, the same answer. Got it? Now put k equals one half into x and y. x is minus one half plus two, over three halves, which gives one. For y, one half times two is one, so y is one plus four, over three halves, which gives ten over three. With me? So the point is one, ten over three, zero. Its z coordinate is zero, so it really sits on the x y plane. One unknown, one condition, and that's it!

5:12Example 2: image of (1, 0, 0) in a line, then a quick check

Now a harder one, Example two, problem seven of Set one, marked medium, and it is exactly our opening challenge. Find the image of the point one, zero, zero in the line x over one equals y minus one over two equals z minus two over three. Start your clock, pause the video, and try it first. Done? Okay, let's check. Our first move is the general point. x over one means x minus zero, over one, so the line passes through zero, one, two, that is vector a, and its direction, vector b, is one, two, three, from the denominators. So N is t, then one plus two times t, then two plus three times t. Okay? Now subtract P, which is one, zero, zero. N minus P is t minus one, then one plus two times t, then two plus three times t. Right? Now the one condition. This vector, dotted with one, two, three, must equal what? Zero, yes, because of the right angle! So the first bracket, plus two times the second, plus three times the third, equals zero. Collect the terms, and fourteen times t, plus seven, equals zero. So t equals minus one half. Clear? Put t equals minus one half into the general point. x is minus one half; y is one minus one, which gives zero; and z is two minus three halves, which gives one half. So the foot N is minus one half, zero, one half. Got it? Now the image, and this is where many students slip. Is it two P minus N, or two N minus P? It is two N minus P, because N is the midpoint of P and P prime. Two N is minus one, zero, one. Subtract one, zero, zero, and the image is minus two, zero, one. With me? And now a ten second check. P prime minus P is minus three, zero, one. Dot it with one, two, three: minus three, plus zero, plus three, which gives zero. So P P prime really is perpendicular to the line. Our challenge is solved, well within two minutes!

7:23Quick revision: section ratio, foot, image 2N minus P

So, quick revision. Section formula: m goes with the far end, and the x y plane divides a join in the ratio minus z one to z two. Foot: write the general point, vector a plus t times vector b, then make N minus P, dot vector b, equal to zero. Image: two N minus P, then check that P P prime is perpendicular to vector b. Now go and try the practice problem on this page. All the best!