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teyona928
Group Title
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
 one year ago
 one year ago
teyona928 Group Title
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
 one year ago
 one year ago

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across Group TitleBest 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.
 one year ago

across Group TitleBest ResponseYou've already chosen the best response.0
I'll do (a) for you:\[ \mathbf{A}\cdot\mathbf{B}=5\cdot5+(9)\cdot(5)=20,\\ \mathbf{A}=\sqrt{5^2+(9)^2}=\sqrt{106},\\ \mathbf{B}=\sqrt{5^2+(5)^2}=5\sqrt{2},\\ \cos^{1}\left(\frac{\mathbf{A}\cdot\mathbf{B}}{\mathbf{A}\mathbf{B}}\right)=\cos^{1}\left(\frac{2}{\sqrt{53}}\right)=105.9^\circ. \]
 one year ago

teyona928 Group TitleBest ResponseYou've already chosen the best response.0
how did you get sqrt of 53
 one year ago

across Group TitleBest ResponseYou've already chosen the best response.0
\[5\sqrt{2}\sqrt{106}=5\sqrt{212}=5\cdot2\sqrt{53}=10\sqrt{53}.\]
 one year ago

teyona928 Group TitleBest ResponseYou've already chosen the best response.0
for a the answer is actually 15.9
 one year ago

across Group TitleBest ResponseYou've already chosen the best response.0
That's right; you have to subtract \(90^\circ\) to find the smallest angle.
 one year ago

oldrin.bataku Group TitleBest 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!
 one year ago

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

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

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.0
dw:1371631085843:dw
 one year ago
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