Expressing a Vector in a new Basis

Question

This is the first time I have attempted such a problem, and have been tackling it all morning. I've been trying to use simultaneous equations to solve for the variables, but it doesn't seem to be working out, so perhaps that's not the way to go...

$$\hat{\bf a} = \dfrac{1}{\sqrt{14}}({\bf i}+2{\bf j}+3{\bf k})$$ $$\hat{\bf b} = \dfrac{1}{\sqrt{5}}(2{\bf i}-{\bf j})$$ $$\hat{\bf c} = \dfrac{1}{\sqrt{70}}(3{\bf i}+6{\bf j}-5{\bf k})$$

Express the vector ${\bf d} = 5{\bf i}+3{\bf j}+2{\bf k}$ in the new $(\hat{\bf a},\hat{\bf b},\hat{\bf c})$ basis.

My attempted solution can be seen here: https://drive.google.com/file/d/0B8CnEriPXOd1T3Z1amU4UGlMVU0/view

There is basic arithmetic mistake in my attempted answer, so I have answered my own question below. Sorry for wasting anybody's time; I'll try to be more careful.

Answer 1

You're looking for $[r,s,t]$ where $r\hat{\bf a}+s\hat{\bf b}+t\hat{\bf c}=5{\bf i}+3{\bf j}+2{\bf k}$.

This means that $\dfrac{r}{\sqrt{14}}+\dfrac{2s}{\sqrt{5}}+\dfrac{3t}{\sqrt{70}}=5$ (the coefficients of ${\bf i}$). Same for the coefficients of ${\bf j}$ and ${\bf k}$. Then it's just a matter of slugging through this rather unattractive system of equations. :(

Answer 2

we will first deal with expressing as a linear combination of $\sqrt{14}\hat a, \sqrt 5\hat b,$ and $\sqrt{70}\hat c.$
we need number $x, y, z$ so that $$x(i+2j+3k)+y(2i-j)+z(3i+6j-5k)=5i+3j+2k \tag 1 $$ in matrix form, we have $$\pmatrix{1&2&3&|&5\\2&-1&6&|&3\\3&0&-5&|&2}\to \pmatrix{1&0&0&|&1.21428\\0&1&0&|&1.40000\\0&0&1&|&0.32857}$$ the solutions $\pmatrix{x\\y\\z}$ is the last column in the second matrix. now, multiplying $x,y,z$ by $\sqrt{14}, \sqrt{5}, \sqrt{70}$ gives you the answer.

Answer 3

Wow, after all of that, and what I had believed to be a meticulous examination of my work, I rechecked it in the morning after some refreshing sleep, and found a basic arithmetic mistake. Theta should have been (17 x rt(14))/14. Sorry to have wasted anybody's time!

Answer 4

so you are looking for [$\alpha, \beta, \gamma$] where $\vec b = \alpha \hat a + \beta \hat b + \gamma \hat c =5\hat i+3\hat j+2\hat k$.

By rules of vectors, if you project $\vec b$ onto $\hat i$, you get the value of that unit vector only, which would be 5 in this case.

mathmatically it looks like this

$\vec d ⋅\hat i = ( 5\hat i\hat i ) + ( 2\hat j\hat i) + ( 3\hat k\hat i )$

where $\hat i\hat i = 1$ and $ \hat j\hat i = \hat k\hat i = 0$

so, $\vec d ⋅\hat i = 5 \times 1 = 5$

we can then do the same with each unit vector you have in your new basis

$\vec d ⋅\hat a = \frac{1}{\sqrt14}(5\times1 + 3\times2 + 2\times3) =\frac{17}{\sqrt14} = \alpha$

$\vec d ⋅\hat b = \frac{1}{\sqrt5}(5\times2 + 3\times-1 + 2\times0) =\frac{7}{\sqrt5} = \beta$

$\vec d ⋅\hat c = \frac{1}{\sqrt70}(5\times3 + 3\times6 + 2\times-5) = \frac{23}{\sqrt70} = \gamma$

Finally we can say that vector $\vec b$ with unit basis $(\hat a, \hat b, \hat c)$ is equal to

$\vec d = \frac{17}{\sqrt14} \hat a + \frac{7}{\sqrt5} \hat b + \frac{23}{\sqrt70} \hat c$