Multiple-choice questions are amenable to a new set of tools that are not available otherwise. To solve a multiple-choice question, you just need to figure out which option is correct. You do not necessarily need to solve the question itself.

As JEE 2015 approaches fast, let me demonstrate this method by showing you how to solve a question from JEE 2012 without really solving it.

Q. Perpendiculars are drawn from points on the line \(\frac{x+2}{2} = \frac{y+1}{-1} = \frac{z}{3}\) to the plane \(x+y+z=3\). The feet of the perpendiculars lie on the line:

- \(\frac{x}{5} = \frac{y-1}{8} = \frac{z-2}{-13}\)
- \(\frac{x}{2} = \frac{y-1}{3} = \frac{z-2}{-5}\)
- \(\frac{x}{4} = \frac{y-1}{3} = \frac{z-2}{-7}\)
- \(\frac{x}{2} = \frac{y-1}{-7} = \frac{z-2}{5}\)

Before starting to solve the question, look at the four options. All of them are of the form \(\frac{x}{a} = \frac{y-1}{b} = \frac{z-2}{c}\). What does this mean?

To specify a line in three (or higher) dimensions, you need to specify two things: its slope and one point that it passes through. In the representation above, the numerators are used to specify the latter and the denominators are used to specify the former. In particular, note that the point \((0, 1, 2)\) lies on the line in all four options above, since if we put \(x = 0, y = 1, z = 2\) in the four equations above, we get \(0 = 0 = 0\) in all four of them. This means the question is not really asking you to find the line defined by the feet of the perpendiculars. It’s asking you to only find the *slope* of the line defined by the feet of the perpendiculars. So if you solve the entire question, you will end up doing extra work—work that you were not even asked to do!

So how do you calculate the slope of the line? Before starting to frantically scribble formulas on a piece of paper, let’s do a few preliminary tests. What properties should the slope of the line we are looking for satisfy? We are looking for the line that is defined by the feet of the perpendiculars drawn from a given line to a given plane. The foot is where the perpendicular intersects the plane. This means the foot always lies on the given plane and that means the line we are looking for must also lie in the given plane. What property does the slope of a line lying in the plane satisfy? Yes, the slope must be perpendicular to any line that’s perpendicular to the plane.

For convenience, let’s call the given line \(L\), the given plane \(P\), and the line we need to find \(L’\). Lines perpendicular to \(P\) are all parallel to the vector \((1, 1, 1)\). That means the slope of \(L’\) should be such that \(L’\) is perpendicular to the vector \((1, 1, 1)\). So if \(L’\) is \(\frac{x}{a} = \frac{y-1}{b} = \frac{z-2}{c}\), then the inner product between \((1, 1, 1)\) and \((a, b, c)\) must be 0, that is, $a+b+c = 0$. So that’s a good test. We can now immediately eliminate all the options that do not satisfy this property. Unfortunately, there is no such option—all of them satisfy the property. The examiner was clever!

Another nice thing to test is whether \(L\) intersects \(P\) or not. If it doesn’t, then \(L\) is parallel to \(P\) and that will imply that \(L’\) must have the same slope as \(L\). That will immediately give us the answer we are looking for. If \(L\) is not parallel to \(P\), however, then it must intersect \(P\). Let’s say the intersection point is \(p\). Where is the foot of the perpendicular drawn from \(p\)? Yes, the foot is also at \(p\). That means \(p\) must lie on \(L’\) as well. Knowing a point that lies on \(L’\) is only going to help us find \(L’\). So it seems this is going to be a useful test.

The easiest way to check whether \(L\) intersects \(P\) or not is to see if it’s parallel to \(P\) or not. It is parallel to \(P\) only if the inner product of \((2, -1, 3)\) with \((1, 1, 1)\) is 0. But the inner product is 4, which means \(L\) intersects \(P\). Good, that means if we find the intersection point \(p\), we will get a point that lies on \(L’\). But there is another issue. We already know a point that lies on \(L’\), namely, the point \((0, 1, 2)\). We know this just because all four options pass through \((0, 1, 2)\). So what if this is the point \(p\)? In that case, we will not gain any new information. So it’s worthwile to check if \(p\) is the point \((0, 1, 2)\) or not. The easiest way to check that is to simply substitute \((0, 1, 2)\) into the equation of \(L\) and see if it satisfies. It does not. So \((0, 1, 2)\) does not lie on \(L\) and hence cannot be \(p\). This is great news! This means all we need to do is find \(p\), because that will give us not one, but *two* points that lie on \(L’\)! And two points uniquely determine a line!

So let’s find \(p\). This is the only time we will do some real calculations. Everything so far was just either trivial substitutions or spatial reasoning coupled with good imagination. To find where \(L\) intersects \(P\), we write \(L\) in a parametric form. That is, we write \(\frac{x+2}{2} = \frac{y+1}{-1} = \frac{z}{3} = t\) and conclude that any point lying on \(L\) must be of the form \((2t-2, -t-1, 3t)\) for some value of \(t\). Now substituting this into the equation for \(P\), we find that \(t = 1.5\) and therefore \(p=(1, -2.5, 4.5)\). Now the other point that lies on \(L’\) is \((0,1,2)\) and therefore the slope of \(L’\) is \((1, -3.5, 2.5)\), or renormalizing, we get a slope of \((2, -7, 5)\). And now we are done. Correct answer is option 4.

To summarize, here is what we did. Let \(L\) and \(P\) be the line and the plane given in the question, and \(L’\) be the line we need to find.

- Realize that \(L’\) must pass through \((0,1,2)\). Why? Because all four options pass through \((0,1,2)\). Note that this is something you would have missed if you had started solving the question without looking at the options.
- Notice that \(L\) is not parallel to \(P\) and thus it must intersect \(P\). Let’s call the intersection \(p\).
- Notice that \((0,1,2)\) does not lie on \(L\). Thus if we find \(p\), we will get
*two*points that pass through \(L’\) thus uniquely determining \(L’\). - Find \(p\), subtract it from \((0,1,2)\) to calculate the slope of \(L’\). Pick option 4 because that has the correct slope.

I know what you are thinking at this point. “Where do I find more tips and tricks like this one?” To the best of our knowledge, we do not really know of a resource that demonstrates these tricks catered towards the IIT-JEE. However, the book titled Street-Fighting Mathematics written by the MIT professor Sanjoy Mahajan does attempt to teach the art of “educated guessing and opportunistic problem solving” and we highly recommend it. The book is also accompanied by an online course. More information can be found here.

*Friday, May 8th, 2015*

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