Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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)).
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
