Calculus Homework Help two

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menal

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how would you do:

1. the vectors x and y are each of length 3 units if x+y=sqrt17 determine x-y
*do you use the cosine law to figure out the angle of x+y (39.5 degrees) and then sub it in the opposite cosine law for x-y to find x-y?

2.show that the vector a=(1,12,-29) can be written as a linear combination of b=(3,1,4) and c=(1,2,-3)
I did this question by: m(3,1,4) + n(1,2,-3)
and then expanded (3m+m+4m)+(1n+2n-3n)
and thn got
1. 3m+n=1
2. m+2n=12
3. 4m-3n=-29
what would i do after that?
 
1a) You'd better check that angle again. 3^2 + 3^2 = 18 is just barely greater than 17. That angle had better be a lot greater than 39.5º. An equilateral triangle would give 60º. A little more spread wouldn't shrink!

1b) You have the right idea, but your "opposite cosine law" will need the supplementary angle.

Let's see what you get on the next go-round.
 
this is what i did:
equation: x+y^2=x^2 + y^2 -2(x)(y)cos<ABC
sqrt17=(3)^2 + (3)^2 -2(3)(3)cosA
sqrt17=9+9-18cosA
sqrt17= 18-18cosA
0=18-sqrt17-18cosA
0=13.87-18cosA
18cosA/18=13.87/18
cosA=0.77
cos-1 (inverse)=39.5 degrees. (approx)
 
I see. Let's wok on the left hand side of the original equation.

No: \(\displaystyle \sqrt{17}\)

Yes: \(\displaystyle \left(\sqrt{17}\right)^2 = 17\)

Think about that equilateral triangle some more. \(\displaystyle \sqrt{17} = 4.123 > 3\) This MUST mean that the angle you seek MUST be greater than 60º. If your best formula suggests something smaller, something is wrong. Always have your best estimate in your head.
 
oooh i did that the first time around but then i confused myself.
i did it again and now i got 1/18 which gave me an angle of 86.8 degrees
so now i just plug it in to the cosine function, but because im tryin to find x-y i change up the equation to
x^2-y^2+2(x)(y)cos86.8 right?
 
Not quite. You'll need 180º - 86.8º. You tell me why.
 
to get the remaining angle?
so the answer is 93.2
and i plug that into the equation.
 
You're not going to make me do all the work, are you? In a vector sense, what's the difference between addition and subtraction?
 
Hello, menal!

\(\displaystyle \text{2. Show that the vector }\vec a\,=\,(1,12,\text{-}29)\text{ can be written as a linear combination of }\vec b\,=\,(3,1,4)\text{ and }\vec c\,=\,(1,2,-3)\)

\(\displaystyle \text{I did this question by: }\:m(3,1,4) + n(1,2,-3) \:=\:(1,12,\text{-}29)\)

\(\displaystyle \text{and then expanded: }\: (3m,m,4m)+(1n,2n,-3n) \:=\:(1,12,\text{-}29)\)

\(\displaystyle \text{and then got: }\;\begin{Bmatrix}1. & 3m+n&=&1 \\ 2.& m+2n&=&12 \\ 3.& 4m-3n&=&\text{-}29 \end{Bmatrix}\) . . All correct! . . . Good work!

\(\displaystyle \text{What would i do after that?}\)
Click to expand...

\(\displaystyle \text{Solve the system of equations and show that it has a unique solution.}\)


\(\displaystyle \text{You'll find that }m = \text{-}2,\;n = 7\text{ satisfies all }three\text{ equations.}\)

\(\displaystyle \text{Therefore: }\:\vec a \;=\;\text{-}2\vec b + 7\vec c\)

 
Hello, menal!

Did you make a sketch?

\(\displaystyle \text{1. The vectors }\vec x\text{ and }\vec y\text{ are each of length 3.}\)
\(\displaystyle \text{If }\,|\vec x+\vec y|\,=\,\sqrt{17},\,\text{ determine }|x-y|.\)
Click to expand...
Code: 
              C
              *
       __   *@ *
      /17 *     * 3
        * @      *
    A *  *  *  *  * B
            3

\(\displaystyle \vec x \:=\:\overrightarrow{AB},\;|\vec x| \,=\,3\)
\(\displaystyle \vec y \:=\:\overrightarrow{BC},\;|\vec y| \,=\,3\)

\(\displaystyle \vec x + \vec y \:=\:\overrightarrow{AC},\;\;|\vec x + \vec y| \:=\:\sqrt{17}\)

\(\displaystyle \text{Let }\angle A \,=\,\angle C \,=\,\theta\)


\(\displaystyle \text{Law of Cosines: }\:\cos A \:=\:\frac{3^2 + (\sqrt{17})^2 - 3^2}{2(3)(\sqrt{17})} \:=\;\frac{17}{6\sqrt{17}} \:=\:\frac{\sqrt{17}}{6}\)

\(\displaystyle \text{Then: }\:\sin^2\!A \:=\:1-\cos^2\!A \:=\:1 - \left(\frac{\sqrt{17}}{6}\right)^2 \:=\:1 - \frac{17}{36} \:=\:\frac{19}{36}\)

. . \(\displaystyle \text{Hence: }\:\sin\theta \:=\:\frac{\sqrt{19}}{6}\)


Code: 
             3
    A *  *  *  *  * B
          *     2@ *
              *     * 3
                  *  *
                      *
                      D

\(\displaystyle \overrightarrow{AB} \,=\,\vec x,\;\;\;\overrightarrow{BD} \:=\:-\vec y\)
\(\displaystyle \overrightarrow{AD} \,=\,\vec x - \vec y\)

\(\displaystyle \text{Note that: }\:\angle B \,=\,2\theta \;\;\; (\angle ABD\text{ is an exterior angle to }\Delta ABC.)\)


\(\displaystyle \text{Law of Cosines:}\)

. . \(\displaystyle AD^2 \;=\;3^2 + 3^2 - 2(3)(3)\cos2\theta \;\;=\;\;18 - 18\cos2\theta \;\;=\;\;18(1 - \cos2\theta)\)

. . . . . . \(\displaystyle =\;18\cdot2\left(\frac{1-\cos2\theta}{2}\right) \;\;=\;\;36\sin^2\!\theta\)

. . . \(\displaystyle AD \;=\;6\sin\theta\)


\(\displaystyle \text{Therefore: }\:|\vec x - \vec y| \;=\;AD \;=\;6\left(\frac{\sqrt{19}}{6}\right) \;=\;\sqrt{19}\)

 
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