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

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I assume you know about the dot product, and that for two vectors [tex]\mathbf a[/tex] and [tex]\mathbf b[/tex], the angle between them [tex]\theta[/tex] satisfies[tex]\mathbf a\cdot\mathbf b=\|\mathbf a\|\|\mathbf b\|\cos\theta\iff\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{\|\mathbf a\|\|\mathbf b\|}[/tex]Then the vectors are parallel if the angle between them is 0 or 180 degrees (0 or pi radians), which would make [tex]\cos\theta=1[/tex] or [tex]\cos\theta=-1[/tex], respectively.Part A)[tex]\vec v_1=\langle\sqrt3,1\rangle\implies\|\vec v_1\|=\sqrt{(\sqrt3)^2+1^2}=\sqrt4=2[/tex][tex]\vec v_2=\langle-\sqrt3,-1\rangle=-\vec v_1\implies\|\vec v_2\|=\|\vec v_1\|=2[/tex][tex]\vec v_1\cdot\vec v_2=(\sqrt3)(-\sqrt3)+(1)(-1)=-4[/tex]Then the angle between [tex]\vec v_1,\vec v_2[/tex] is such that[tex]\cos\theta=\dfrac{-4}{(2)(2)}=-1\implies\theta=\pi\,\mathrm{rad}[/tex]so these vectors are parallel ("antiparallel", more specifically, which means they are parallel but point in opposite directions).Part B) involves the same computations:[tex]\vec u_1=\langle2,3\rangle\implies\|\vec u_1\|=\sqrt{2^2+3^2}=\sqrt{13}[/tex][tex]\vec u_2[/tex] has the same components but differing by sign and order, as [tex]\vec u_1[/tex]; its magnitude remains the same, though:[tex]\vec u_2=\langle-3,-2\rangle\implies\|\vec u_2\|=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}[/tex][tex]\vec u_1\cdot\vec u_2=(2)(-3)+(3)(-2)=-12[/tex][tex]\implies\cos\theta=\dfrac{-12}{(\sqrt{13})(\sqrt{13})}=-\dfrac{12}{13}\implies\theta=\cos^{-1}\left(-\dfrac{12}{13}\right)[/tex]which is neither 0 nor pi, which means these vectors are not parallel.