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 + \frac13\,\vec\jmath + \frac13\,\vec k = \left\langle1,\frac13,\frac13\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*(-3)+1*(-1)+(-1)*1 = -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\| = √(3^2 + 1^2 + (-1)^2) = √(11) \\\\ \|\vec b\| = √((-3)^2+(-1)^2+1^2) = √(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,\frac13,\frac13\right\rangle = 3*1+1*\frac13+(-1)*\frac13 = 3 \\\\ \|\vec a\|\|\vec c\| = √(3^2+1^2+(-1)^2)*\sqrt{1^2+\frac1{3^2}+\frac1{3^2}} = \frac{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) = (\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)=\frac3{\frac{11}3}\right) =\frac9{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)).