Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 30

MATH152 April 2011

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Question A 30
For questions A29 and A30 below, decide if the statements are true or false. In each case give brief justification of your conclusion. If vectors v and w in $\mathbb {R} ^{3}$ are linearly independent and lie in the plane P that goes through the origin then $\mathbf {v} \times (\mathbf {v} \times \mathbf {w} )$ also lies in P.

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
How do vectors in the plane relate to the vector $v\times {w}$?

Hint 2
If a vector y is in the plane then it must be written as a linear combination of v and w since they span the plane.

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. The normal of the plane is given by, $\mathbf {n} =\mathbf {v} \times \mathbf {w}$ and therefore if a vector is in the plane it must be orthogonal to this normal vector. Recall that vectors are orthogonal if the scalar (dot) product is zero. Therefore we seek to check if, $(\mathbf {v} \times (\mathbf {v} \times \mathbf {w} ))\cdot (\mathbf {v} \times \mathbf {w} )=(\mathbf {v} \times \mathbf {n} )\cdot \mathbf {n} =0.$ To do this we use the triple scalar triple product which for any three vectors a, b, and c states, $\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )=\mathbf {c} \cdot (\mathbf {a} \times \mathbf {b} )=\mathbf {b} \cdot (\mathbf {c} \times \mathbf {a} ).$ Therefore we see that, $\mathbf {n} \cdot (\mathbf {v} \times \mathbf {n} )=\mathbf {v} \cdot (\mathbf {n} \times \mathbf {n} )=0$ by the virtue that parallel vectors have a zero cross-product. Therefore we see that the vector $\mathbf {v} \times (\mathbf {v} \times \mathbf {w} )$ is orthogonal to the normal vector of the plane and thus lies in the plane itself. Therefore, the statement is true.

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. From hint 2, we recall that since the plane is in $\mathbb {R} ^{2}$, the two linearly independent vectors inside are enough to span the entire space. Therefore, any vector, y, inside P can be written as a linearly dependent combination of v and w, $\mathbf {y} =d_{1}\mathbf {v} +d_{2}\mathbf {w}$ for some scalars, $d_{1}$ and $d_{2}$. Recall that for any vectors, a, b, and c, we can define the vector triple product as $\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} .$ We wish to see if the vector $\mathbf {v} \times (\mathbf {v} \times \mathbf {w} )$ is inside the plane P. Using the vector triple product, $\mathbf {v} \times (\mathbf {v} \times \mathbf {w} )=(\mathbf {v} \cdot \mathbf {w} )\mathbf {v} -(\mathbf {v} \cdot \mathbf {v} )\mathbf {w} .$ But this is a linear combination of the vectors $\mathbf {v}$ and $\mathbf {w}$, and hence is inside the plane P. To see this more clearly, define ${\begin{aligned}d_{1}&=(\mathbf {v} \cdot \mathbf {w} )\\d_{2}&=-(\mathbf {v} \cdot \mathbf {v} )\end{aligned}}$ then we have that $\mathbf {v} \times (\mathbf {v} \times \mathbf {w} )=d_{1}\mathbf {v} +d_{2}\mathbf {w}$ and therefore the vector $\mathbf {v} \times (\mathbf {v} \times \mathbf {w} )$ is inside the plane P, making the statement true.