Question

onsider the vectors a→=3i→+j→−k→,b→=−3i→−j→+k→,c→=i→+0.333333j→+0.333333k→ d→=i→+3j→+4k→,g→=i→+3j→−k→. Which pairs (if any) of these vectors are (a) Are perpendicular? (Enter none or a pair or list of pairs, e.g., if a→ is perpendicular to b→ and c→, enter (a,b),(a,c).) (b) Are parallel? (Enter none or a pair or list of pairs, e.g., if a→ is parallel to b→ and c→, enter (a,b),(a,c).) (c) Have an angles less than π/2 between them? (Enter none or a pair or list of pairs, e.g., if a→ is at an angle less than pi/2 from b→ and c→, enter (a,b),(a,c).) (d) Have an angle of more than π/2 between them? (Enter none or a pair or list of pairs, e.g., if a→ is at an angle greater than pi/2 from b→ and c→, enter (a,b),(a,c).)

Answer 1

We're given the vectors

$\vec a = 3\vec\imath + \vec\jmath - \vec k = \langle3,1,-1\rangle \\\\ \vec b = -3\,\vec\imath-\vec\jmath + \vec k = \langle-3,-1,1\rangle \\\\ \vec c = \vec\imath + \dfrac13\,\vec\jmath + \dfrac13\,\vec k = \left\langle1,\dfrac13,\dfrac13\right\rangle \\\\ \vec d = \vec\imath + 3\,\vec\jmath + 4\,\vec k = \langle 1,3,4\rangle \\\\ \vec g = \vec\imath + 3\,\vec\jmath - \vec k = \langle1,3,-1\rangle$

(a) Two vectors are perpendicular if their dot product is zero. For instance, $\vec a$ and $\vec b$ are not perpendicular because

$\vec a\cdot\vec b = \langle3,1,-1\rangle\cdot\langle-3,-1,1\rangle = 3\times(-3)+1\times(-1)+(-1)\times1 = -11$

You'll find that none of these vectors taken two at a time are perpendicular to each other.

(b) Recall for any two vectors $\vec x$ and $\vec y$ that

$\vec x\cdot\vec y = \|\vec x\| \|\vec y\| \cos(\theta)$

where $\theta$ is the angle between $\vec x$ and $\vec y$. 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

$\vec x\cdot\vec y = \pm\|\vec x\| \|\vec y\|$

For instance, we know that

$\vec a\cdot\vec b = -11$

and we have

$\|\vec a\| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11} \\\\ \|\vec b\| = \sqrt{(-3)^2+(-1)^2+1^2} = \sqrt{11}$

so $\vec a$ and $\vec b$ are indeed parallel and point in opposite directions, since -11 = - √11 × √11.

On the other hand, $\vec a$ and $\vec c$ are not parallel, since

$\vec a\cdot\vec c = \langle3,1,-1\rangle\cdot\left\langle1,\dfrac13,\dfrac13\right\rangle = 3\times1+1\times\dfrac13+(-1)\times\dfrac13 = 3 \\\\ \|\vec a\|\|\vec c\| = \sqrt{3^2+1^2+(-1)^2}\times\sqrt{1^2+\dfrac1{3^2}+\dfrac1{3^2}} = \dfrac{11}3$

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

$\cos(\theta) = \dfrac{\vec x\cdot\vec y}{\|\vec x\|\|\vec y\|}$

For instance, we know $\vec a$ and $\vec b$ are parallel and have an angle of π rad between them. cos(π) = -1, so this pair doesn't qualify. Meanwhile, the angle between

$\cos(\theta)=\dfrac3{\frac{11}3}\right) =\dfrac9{11} > 0$

so $\vec a$ and $\vec c$ 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)).