Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 28

MATH152 April 2017

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[hide] Question A 28
Let P be the plane in $\mathbb {R} ^{3}$ with equation $x+y-z=0$. Find two more planes $Q$ and $R$ such that all three planes are different and the intersection is the line $s[3,-2,1]+[2,2,4]$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
Think carefully about the geometry of this problem. A plane will contain a line L whenever the plane contains a point on L and the normal vector to the plane is perpendicular to L. So we want to find a plane P whose normal vector is perpendicular to the given line, such that P contains the point (2,2,4). One possible normal vector is (1, 1, -1). Just find 2 others.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

[show] Solution

A plane will contain a line L whenever the plane contains a point on L and the normal vector to the plane is perpendicular to L. So we want to find a plane P whose normal vector is perpendicular to the given line, such that P contains the point (2,2,4). We need to find vectors perpendicular to the vector [3, -2, 1]. In order to guarantee that all 3 planes are different, we will pick the first two components of the normal to Q to be 1 and 2, and the first two components of the normal to R to be 1 and 3. (There are many solutions to this problem. You may prefer to use different normal vectors than these). Under this simplifying assumption, the normal to Q is of the form [1,2,q] and the normal to R is of the form [1,3,r]. The dot product of [1,2,q] and [3,-2,1] is $3-4+q=q-1$. This is equal to zero when q = 1. Similarly, the dot product of [3, -2, 1] and [1,3,r] is $3-6+r=r-3$. This is equal to zero when r = 3. So the normal vector to Q is [1,2,1] and the normal vector to R is [1,3,3]. Beecause both Q and R should contain the point (2,2,4), we get the equation $1(x-2)+2(y-2)+1(z-4)=0$ for Q, and $1(x-2)+3(y-2)+3(z-4)=0$ for R. One Possible Answer: $\color {blue}(x-2)+2(y-2)+(z-4)=0;(x-2)+3(y-2)+3(z-4)=0$