Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 11

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MATH152 April 2017

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Question A 11
Let $a=[0,1,1]$ and $b=[2,2,0]$ . Sketch the set of points $\{sa+(1-s)b:0\leq s\leq 1\}$ on the graph below.

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!

Hint
Observe that $sa+(1-s)b$ can be written as $b+s(a-b)$ . What point is this when $s=0$ ? How about when $s=1$ ? And when $0<s<1$ ?

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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.
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Solution
The location of $sa+(1-s)b=b+s(a-b)$ can be thought of as the endpoint of the vector sum of ${\vec {b}}$ and $s({\vec {a}}-{\vec {b}})$ as shown below. Notice that $s({\vec {a}}-{\vec {b}})$ is a scaled version of ${\vec {a}}-{\vec {b}}$ , and since $0\leq s\leq 1$ , its length ranges between 0 and the full length of ${\vec {a}}-{\vec {b}}$ . MATH152 April 2017 Question A11 Hence the set in question is simply a line segment between ${\vec {b}}$ and ${\vec {a}}$ : MATH152 April 2017 Question A11 (2)