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teyona928

help plz need help immediately!! Using the definition of the scalar product, find the angle between the following vectors. (Find the smallest angle.) a) A = 5i - 9j and B = 5i - 5j (b) A = -8i + 5j and B = 3i - 4j + 2k (c) A= -2i + 2j and B = 3j + 4k

  • 10 months ago
  • 10 months ago

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  1. across
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    The scalar product is defined as follows:\[ \mathbf{A}\cdot\mathbf{B}=|\mathbf{A}||\mathbf{B}|\cos\theta. \]Therefore,\[ \theta=\cos^{-1}\left(\frac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{A}||\mathbf{B}|}\right). \]Then use the fact that\[ \mathbf{A}\cdot\mathbf{B}=\sum a_ib_i=\mathbf{A}\mathbf{B}^T \]to wrap it up.

    • 10 months ago
  2. across
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    I'll do (a) for you:\[ \mathbf{A}\cdot\mathbf{B}=5\cdot5+(-9)\cdot(-5)=-20,\\ |\mathbf{A}|=\sqrt{5^2+(-9)^2}=\sqrt{106},\\ |\mathbf{B}|=\sqrt{5^2+(-5)^2}=5\sqrt{2},\\ \cos^{-1}\left(\frac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{A}||\mathbf{B}|}\right)=\cos^{-1}\left(-\frac{2}{\sqrt{53}}\right)=105.9^\circ. \]

    • 10 months ago
  3. teyona928
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    how did you get sqrt of 53

    • 10 months ago
  4. across
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    \[5\sqrt{2}\sqrt{106}=5\sqrt{212}=5\cdot2\sqrt{53}=10\sqrt{53}.\]

    • 10 months ago
  5. teyona928
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    for a the answer is actually 15.9

    • 10 months ago
  6. across
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    That's right; you have to subtract \(90^\circ\) to find the smallest angle.

    • 10 months ago
  7. oldrin.bataku
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    NO!

    • 10 months ago
  8. oldrin.bataku
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    @across that's incorrect; you computed the dot product incorrectly so your angle is actually between \((5,-9)\) and \(5,5\) -- notice one of our vectors now points \(90^\circ\) away from where it once did!

    • 10 months ago
  9. oldrin.bataku
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    $$\vec{a}=5\mathbf{i}-9\mathbf{j}\\\vec{b}=5\mathbf{i}-5\mathbf{j}\\\vec{a}\cdot\vec{b}=5(5)+(-9)(-5)=25+45=70$$

    • 10 months ago
  10. oldrin.bataku
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    $$\|\vec{a}\|=\sqrt{5^2+(-9)^2}=\sqrt{25+81}=\sqrt{106}\\\|\vec{b}\|=\sqrt{5^2+(-5)^2}\sqrt{2(5^2)}5=5\sqrt2\\\cos \theta=\frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\|\vec{b}\|}=\frac{70}{5\sqrt2\sqrt{106}}=\frac{14}{2\sqrt{53}}=\frac7{\sqrt{53}}\implies\theta=\arccos\frac7{\sqrt53}\approx15.9^\circ$$

    • 10 months ago
  11. oldrin.bataku
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    |dw:1371631085843:dw|

    • 10 months ago
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