## teyona928 2 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

1. across

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.

2. across

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

3. teyona928

how did you get sqrt of 53

4. across

$5\sqrt{2}\sqrt{106}=5\sqrt{212}=5\cdot2\sqrt{53}=10\sqrt{53}.$

5. teyona928

for a the answer is actually 15.9

6. across

That's right; you have to subtract $$90^\circ$$ to find the smallest angle.

7. oldrin.bataku

NO!

8. oldrin.bataku

@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!

9. oldrin.bataku

$$\vec{a}=5\mathbf{i}-9\mathbf{j}\\\vec{b}=5\mathbf{i}-5\mathbf{j}\\\vec{a}\cdot\vec{b}=5(5)+(-9)(-5)=25+45=70$$

10. oldrin.bataku

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

11. oldrin.bataku

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