Question

*calculus 3* Let vector U = 4i - 2k and let vector V = i-2j+k
_Find the magnitude (norm) of 3U - 2V - aka II 3U - 2V II
_Two vectors of length 2 which are also parallel to 3U - 2V
_The components and projections of 3U along 2V

Answer 1

hint:
\[3U - 2V = \left( {\begin{array}{*{20}{c}}
{12}&0&{ - 6}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
2&{ - 4}&2
\end{array}} \right)\]

Answer 2

I understand that part. i calculated the magnitude and got 180^1/2
However, i am not sure how to solve the second or third part.

Answer 3

We have:
\[\begin{gathered}
W = 3U - 2V = \left( {\begin{array}{*{20}{c}}
{12}&0&{ - 6}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
2&{ - 4}&2
\end{array}} \right) \hfill \\
W = \left( {\begin{array}{*{20}{c}}
{10}&4&{ - 8}
\end{array}} \right) \hfill \\
\end{gathered} \]

Answer 4

a vector parallel to vector W is the subsequent vector:
\[kW = \left( {\begin{array}{*{20}{c}}
{10k,}&{4k,}&{ - 8k}
\end{array}} \right)\]
so we have to determine the value of k such that the length of kW is equal to 2

Answer 5

Would k= (2/(180^1/2))? I first made the magnitude equal to 1 by dividing it by 180^1/2, then I multiplied by 2 to get me a vector the length of 2.

Answer 6

we have 2 values of k since we can write:
\[\left\| W \right\| = \left| k \right|\sqrt {180} \]
so:
\[k = \pm \frac{2}{{\sqrt {180} }}\]

Answer 7

oops..
\[\left\| {kW} \right\| = \left| k \right|\sqrt {180} \]

Answer 8

In order to get the projection and component we have to compute the Fourier's coefficient of one vector with respect to the other

Answer 9

for example the projection of the vector 3U along the vector 2V is given by the subsequent formula:
\[c2V = \frac{{\left\langle {3U,2V} \right\rangle }}{{\left\langle {2V,2V} \right\rangle }}2V\]

Answer 10

where:
\[\left\langle { \cdot , \cdot } \right\rangle \]
is the scalar product
and c is the Fourier's coefficient:
\[c = \frac{{\left\langle {3U,2V} \right\rangle }}{{\left\langle {2V,2V} \right\rangle }}\]

Answer 11

so we have to compute these quantities:
\[{\left\langle {3U,2V} \right\rangle }\] and
\[{\left\langle {2V,2V} \right\rangle }\]

Answer 12

So I would be doing the dot product for the numerator (in this case 3U and 2V) and for the denominator (in this case 2V and 2V)?

Answer 13

yes! that's the coefficient c, then the projection is the vector:
c 2V

Answer 14

Ah ok, got it! Just hypothetically, if they wanted the reverse "find the projection of 2V along 3U - the Fourier coefficient would be

\[c = \frac{ <2U, 3V> }{<3V,3V> }\]

Answer 15

then I would multiply the coefficent by vector 3U

Answer 16

oops i mixed up the variables for the fraction: 2V and 3U

Answer 17

the projection of the vector 2V along the vector 3U is:
\[{c_1}3U = \frac{{\left\langle {2V,3U} \right\rangle }}{{\left\langle {3U,3U} \right\rangle }}3U\]

Answer 18

Ok good! Thanks for the help! :)

Answer 19

we have to swap 2V with 3U

Answer 20

Thanks! :)