A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years 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
anonymous
 3 years 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

This Question is Closed

across
 3 years ago
Best ResponseYou've already chosen the best response.0The 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how did you get sqrt of 53

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for a the answer is actually 15.9

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

anonymous
 3 years ago
Best 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0$$\vec{a}=5\mathbf{i}9\mathbf{j}\\\vec{b}=5\mathbf{i}5\mathbf{j}\\\vec{a}\cdot\vec{b}=5(5)+(9)(5)=25+45=70$$

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1371631085843:dw
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.