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Sauron [17]
3 years ago
9

HELP PICS INCLUDED! WILL GIVE BRAINLIEST! SHOW WORK AND EXPLAIN

Mathematics
1 answer:
Naddik [55]3 years ago
4 0

I assume you know about the dot product, and that for two vectors \mathbf a and \mathbf b, the angle between them \theta satisfies

\mathbf a\cdot\mathbf b=\|\mathbf a\|\|\mathbf b\|\cos\theta\iff\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{\|\mathbf a\|\|\mathbf b\|}

Then the vectors are parallel if the angle between them is 0 or 180 degrees (0 or pi radians), which would make \cos\theta=1 or \cos\theta=-1, respectively.

Part A)

\vec v_1=\langle\sqrt3,1\rangle\implies\|\vec v_1\|=\sqrt{(\sqrt3)^2+1^2}=\sqrt4=2

\vec v_2=\langle-\sqrt3,-1\rangle=-\vec v_1\implies\|\vec v_2\|=\|\vec v_1\|=2

\vec v_1\cdot\vec v_2=(\sqrt3)(-\sqrt3)+(1)(-1)=-4

Then the angle between \vec v_1,\vec v_2 is such that

\cos\theta=\dfrac{-4}{(2)(2)}=-1\implies\theta=\pi\,\mathrm{rad}

so these vectors are parallel ("antiparallel", more specifically, which means they are parallel but point in opposite directions).

Part B) involves the same computations:

\vec u_1=\langle2,3\rangle\implies\|\vec u_1\|=\sqrt{2^2+3^2}=\sqrt{13}

\vec u_2 has the same components but differing by sign and order, as \vec u_1; its magnitude remains the same, though:

\vec u_2=\langle-3,-2\rangle\implies\|\vec u_2\|=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}

\vec u_1\cdot\vec u_2=(2)(-3)+(3)(-2)=-12

\implies\cos\theta=\dfrac{-12}{(\sqrt{13})(\sqrt{13})}=-\dfrac{12}{13}\implies\theta=\cos^{-1}\left(-\dfrac{12}{13}\right)

which is neither 0 nor pi, which means these vectors are not parallel.

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