Question

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

Answer 1

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.

Answer 2

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

Answer 3

how did you get sqrt of 53

Answer 4

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

Answer 5

for a the answer is actually 15.9

Answer 6

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

Answer 7

NO!

Answer 8

@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!

Answer 9

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

Answer 10

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