Science:Math Exam Resources/Courses/MATH152/April 2022/Question A08

MATH152 April 2022

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Question A08
Let $W$ be the plane in $\mathbb {R} ^{3}$ through the points $$(1,1,0),\ (2,-1,1),\ (0,1,c).$$ For what value (or values) of $c$ does the plane pass through the point $(1,-1,2)$?

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Can you write down an equation that the points on the plane need to satisfy? Note that it will depend on $c$.

Hint 2
For a more slick solution, note that if four points are co-planar, then the three vectors that go from one point to each of the other three must necessarily be linearly dependent.

Checking a solution serves two purposes: helping you if, after having used all the hints, 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.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We will use the following approach. Note that the difference between two points on a plane is a vector that is parallel to the plane. If we can find two such vectors (that are not co-linear), then their cross-product is normal to the plane and can be used to write down an equation that the points on the plane need to satisfy. Let then $${\begin{aligned}{\vec {v}}&=(2,-1,1)-(1,1,0)=&\left[{\begin{array}{c}1\\-2\\1\end{array}}\right]\\{\vec {w}}&=(1,1,0)-(0,1,c)=&\left[{\begin{array}{c}1\\0\\-c\end{array}}\right].\end{aligned}}$$ Their cross-product is $${\vec {v}}\times {\vec {w}}=\left[{\begin{array}{c}2c\\1+c\\2\end{array}}\right].$$ Therefore, a point $(x,y,z)$ in the plane is characterized by the property that the vector from $(1,1,0)$ (or any other point on the plane) to $(x,y,z)$ is orthogonal to ${\vec {v}}\times {\vec {w}}$: $$2c(x-1)+(1+c)(y-1)+2z=0.$$ Substituting $(1,-1,2)$ for $(x,y,z)$ yields the equation $$(1+c)(-2)+4=0,$$ for which $c=1$ is the only solution.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Following the second hint, consider the three vectors from $(1,-1,2)$ to the points $(1,1,0),\ (2,-1,1),\ {\text{ and }}(0,1,c)$. That these three vectors are linearly dependent is characterized by the fact that the parallelepiped that they span has no volume or, equivalently that the matrix formed by these three vectors has determinant equal to 0: $$0=\det \left[{\begin{array}{ccc}0&2&-2\\1&0&-1\\-1&2&c-2\end{array}}\right]=0-2\det \left[{\begin{array}{cc}1&-1\\-1&c-2\end{array}}\right]-2\det \left[{\begin{array}{cc}1&0\\-1&2\end{array}}\right]=-2(c-3)-2(2)=-2c+2.$$ The above equation has unique solution $c=1$.