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Sagot :
We're given the vectors
[tex]\vec a = 3\vec\imath + \vec\jmath - \vec k = \langle3,1,-1\rangle \\\\ \vec b = -3\,\vec\imath-\vec\jmath + \vec k = \langle-3,-1,1\rangle \\\\ \vec c = \vec\imath + \dfrac13\,\vec\jmath + \dfrac13\,\vec k = \left\langle1,\dfrac13,\dfrac13\right\rangle \\\\ \vec d = \vec\imath + 3\,\vec\jmath + 4\,\vec k = \langle 1,3,4\rangle \\\\ \vec g = \vec\imath + 3\,\vec\jmath - \vec k = \langle1,3,-1\rangle[/tex]
(a) Two vectors are perpendicular if their dot product is zero. For instance, [tex]\vec a[/tex] and [tex]\vec b[/tex] are not perpendicular because
[tex]\vec a\cdot\vec b = \langle3,1,-1\rangle\cdot\langle-3,-1,1\rangle = 3\times(-3)+1\times(-1)+(-1)\times1 = -11[/tex]
You'll find that none of these vectors taken two at a time are perpendicular to each other.
(b) Recall for any two vectors [tex]\vec x[/tex] and [tex]\vec y[/tex] that
[tex]\vec x\cdot\vec y = \|\vec x\| \|\vec y\| \cos(\theta)[/tex]
where [tex]\theta[/tex] is the angle between [tex]\vec x[/tex] and [tex]\vec y[/tex]. If these vectors are parallel, then the angle between them is 0 rad or π rad, meaning they point in the same or in opposite directions, respectively.
We have cos(0) = 1 and cos(π) = -1, so
[tex]\vec x\cdot\vec y = \pm\|\vec x\| \|\vec y\|[/tex]
For instance, we know that
[tex]\vec a\cdot\vec b = -11[/tex]
and we have
[tex]\|\vec a\| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11} \\\\ \|\vec b\| = \sqrt{(-3)^2+(-1)^2+1^2} = \sqrt{11}[/tex]
so [tex]\vec a[/tex] and [tex]\vec b[/tex] are indeed parallel and point in opposite directions, since -11 = - √11 × √11.
On the other hand, [tex]\vec a[/tex] and [tex]\vec c[/tex] are not parallel, since
[tex]\vec a\cdot\vec c = \langle3,1,-1\rangle\cdot\left\langle1,\dfrac13,\dfrac13\right\rangle = 3\times1+1\times\dfrac13+(-1)\times\dfrac13 = 3 \\\\ \|\vec a\|\|\vec c\| = \sqrt{3^2+1^2+(-1)^2}\times\sqrt{1^2+\dfrac1{3^2}+\dfrac1{3^2}} = \dfrac{11}3[/tex]
and clearly 3 ≠ ±11/3.
It turns out that (a, b) is the only pair of parallel vectors.
(c) The cosine of an angle measuring between 0 and π/2 rad is positive, so you just need to check the sign of
[tex]\cos(\theta) = \dfrac{\vec x\cdot\vec y}{\|\vec x\|\|\vec y\|}[/tex]
For instance, we know [tex]\vec a[/tex] and [tex]\vec b[/tex] are parallel and have an angle of π rad between them. cos(π) = -1, so this pair doesn't qualify. Meanwhile, the angle between
[tex]\cos(\theta)=\dfrac3{\frac{11}3}\right) =\dfrac9{11} > 0[/tex]
so [tex]\vec a[/tex] and [tex]\vec c[/tex] do qualify.
You'd find that the pairs ((a, c), (a, d), (a, g), (c, d), (c, g), (d, g)).
(d) An angle between π/2 and π has a negative cosine. None of the vectors are perpendicular to each other, so this happens for the remaining pairs, ((a, b), (b, c), (b, d), (b, g)).
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