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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
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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acrossBest ResponseYou've already chosen the best response.0
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

acrossBest ResponseYou've already chosen the best response.0
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

teyona928Best ResponseYou've already chosen the best response.0
how did you get sqrt of 53
 10 months ago

acrossBest ResponseYou've already chosen the best response.0
\[5\sqrt{2}\sqrt{106}=5\sqrt{212}=5\cdot2\sqrt{53}=10\sqrt{53}.\]
 10 months ago

teyona928Best ResponseYou've already chosen the best response.0
for a the answer is actually 15.9
 10 months ago

acrossBest ResponseYou've already chosen the best response.0
That's right; you have to subtract \(90^\circ\) to find the smallest angle.
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
@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

oldrin.batakuBest ResponseYou've already chosen the best response.0
$$\vec{a}=5\mathbf{i}9\mathbf{j}\\\vec{b}=5\mathbf{i}5\mathbf{j}\\\vec{a}\cdot\vec{b}=5(5)+(9)(5)=25+45=70$$
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
$$\\vec{a}\=\sqrt{5^2+(9)^2}=\sqrt{25+81}=\sqrt{106}\\\\vec{b}\=\sqrt{5^2+(5)^2}\sqrt{2(5^2)}5=5\sqrt2\\\cos \theta=\frac{\vec{a}\cdot\vec{b}}{\\vec{a}\\\vec{b}\}=\frac{70}{5\sqrt2\sqrt{106}}=\frac{14}{2\sqrt{53}}=\frac7{\sqrt{53}}\implies\theta=\arccos\frac7{\sqrt53}\approx15.9^\circ$$
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
dw:1371631085843:dw
 10 months ago
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