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3 years ago

10

Solve the equation 11x+16y=13y

1 answer:

aksik [14]3 years ago

6 0

Answer:

x=-3y/11 , y=-11x/3

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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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)

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

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

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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:

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

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

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