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I am trying to solve the following question:

Given that $\overrightarrow{a} = \langle3, 2, -1\rangle$, find a vector $\overrightarrow{b}$ such that $comp_{\overrightarrow{a}}\overrightarrow{b} = 5$

I know the general form of $comp_{\overrightarrow{a}}\overrightarrow{b}$ can be expressed as:

$$comp_{\overrightarrow{a}}\overrightarrow{b} = \frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|}$$

So, we can substitute the given value for $\overrightarrow{a}$ and insert generic variables to denote $\overrightarrow{b}$:

$$\frac{\langle3, 2, -1\rangle\cdot\langle b_1, b_2, b_3\rangle}{\sqrt{3^2 + 2^2 + (-1)^2}} = \frac{3b_1 + 2b_2 -1b_3}{\sqrt{14}} = 5 $$

So simplifying further, we can conclude:

$$3b_1 + 2b_2 -1b_1 = 5\sqrt{14}$$

Clearly, there are an infinite number of vectors that would satisfy this condition.

I want to include sample answers for $\overrightarrow{b}$, but I'm not sure how I can reduce this equation into vector form. How can I represent this generic solution as a vector?

Thanks for you help!

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  • $\begingroup$ You need to read more carefully what the definition of $\text{com}_a(b)$ is. $\endgroup$ – Squirtle Sep 2 '15 at 2:43
  • $\begingroup$ I don't think we're talking about the same thing. I'm referring to $comp_{\overrightarrow{a}}\overrightarrow{b}$, which is the scalar projection. $\endgroup$ – James Taylor Sep 2 '15 at 2:50
  • $\begingroup$ Also, if it helps, this question is very similar in structure to the one posted here. $\endgroup$ – James Taylor Sep 2 '15 at 2:52
  • $\begingroup$ Even according to the question / solution you posted you see that there are many choices for your solution. $\endgroup$ – Squirtle Sep 2 '15 at 2:57
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If you read the question carefully... you will notice it says find "a" vector b such that ... NOT find "the" vector b such that...

The obvious choice is the easiest one, namely:

Let $b=\frac{5}{|a|}a$

Then $$\frac{a\cdot b}{\vert a\vert} =\frac{a\cdot(\frac{5}{|a|}a)}{|a|}= 5\frac{|a|^2}{|a|^2}=5$$

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  • $\begingroup$ So what would $\overrightarrow{b}$ actually be? What are the components? $\endgroup$ – James Taylor Sep 2 '15 at 3:14
  • $\begingroup$ Do you know how scalar multiplication works? $\endgroup$ – Squirtle Sep 2 '15 at 3:17
  • $\begingroup$ No, I just know that this example problem has a bunch of solutions and I have no idea where they're coming from. Similarly, I don't know how to produce them for the problem I worked out. $\endgroup$ – James Taylor Sep 2 '15 at 3:21
  • $\begingroup$ If a vector $x$ is in $\mathbb{R}^3$ then we can label it with components $x=(x_1,x_2,x_3)$. The rule for scalar multiplication by a number $\lambda$ is, $\lambda x = \lambda(x_1,x_2,x_3) = (\lambda x_1,\lambda x_2,\lambda x_3)$. So think about how you can apply this to my solution... what's your $\lambda$ in this case? $\endgroup$ – Squirtle Sep 2 '15 at 3:25
  • $\begingroup$ Given that I have the equation $3b_1 + 2b_2 -1b_1 = 5\sqrt{14}$ to work with, I would assume my $λ$ in this case would be $5\sqrt{14}$ which would be multiplied across every single term. Applying that logic to the example problem fails, however, because the solutions given are not multiples of $2\sqrt{10}$. I know $λ$ cannot be just $5$ because that would defeat the purpose of deriving a general equation. $\endgroup$ – James Taylor Sep 2 '15 at 3:31

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