Question

Consider the points ​P(5​,5​,1​) and ​Q(13​,13​,3​).

a. Find PQ with right arrow and state your answer in two​ forms: (a,b,c) and ai+bj+ck.
b. Find the magnitude of PQ with right arrow.
c. Find two unit vectors parallel to PQ with right arrow.

Answer 1

a)
`\overrightarrow{PQ} = (8,8, 2)` or
`\overrightarrow{PQ} = 8\,i + 8\,j + 2\,k`, b) The magnitude of segment PQ is approximately 11.489, c) The two unit vectors associated to PQ are, respectively:
`\vec v_(1) = (0.696,0.696, 0.174)` and
`\vec v_(2) = (-0.696,-0.696, -0.174)`

Explanation:

a) The vectorial form of segment PQ is determined as follows:

`\overrightarrow {PQ} = \vec Q - \vec P`

Where
`\vec Q` and
`\vec P` are the respective locations of points Q and P with respect to origin. If
`\vec Q = (13,13,3)` and
`\vec P = (5,5,1)`, then:

`\overrightarrow{PQ} = (13,13,3)-(5,5,1)`

`\overrightarrow {PQ} = (13-5, 13-5, 3 - 1)`

`\overrightarrow{PQ} = (8,8, 2)`

Another form of the previous solution is
`\overrightarrow{PQ} = 8\,i + 8\,j + 2\,k`.

b) The magnitude of the segment PQ is determined with the help of Pythagorean Theorem in terms of rectangular components:

`\|\overrightarrow{PQ}\| =\sqrt{PQ_(x)^(2)+PQ_(y)^(2)+PQ_(z)^(2)}`

`\|\overrightarrow{PQ}\| = \sqrt{8^(2)+8^(2)+2^(2)}`

`\|\overrightarrow{PQ}\|\approx 11.489`

The magnitude of segment PQ is approximately 11.489.

c) There are two unit vectors associated to PQ, one parallel and another antiparallel. That is:

`\vec v_(1) = \vec u_(PQ)` (parallel) and
`\vec v_(2) = -\vec u_(PQ)` (antiparallel)

The unit vector is defined by the following equation:

`\vec u_(PQ) = \frac{\overrightarrow{PQ}}{\|\overrightarrow{PQ}\|}`

Given that
`\overrightarrow{PQ} = (8,8, 2)` and
`\|\overrightarrow{PQ}\|\approx 11.489`, the unit vector is:

`\vec u_(PQ) = ((8,8,2))/(11.489)`

`\vec u_(PQ) = \left((8)/(11.489),(8)/(11,489),(2)/(11.489) \right)`

`\vec u_(PQ) = \left(0.696, 0.696,0.174\right)`

The two unit vectors associated to PQ are, respectively:

`\vec v_(1) = (0.696,0.696, 0.174)` and
`\vec v_(2) = (-0.696,-0.696, -0.174)`