Question

Given the vectors M = −10 ax + 4 ay − 8 az and N = 8 ax + 7ay − 2 az, find:

a) a unit vector in the direction of −M +2N,
b) the magnitude of 5 ax + N − 3M, c) |M||2N|(M+ N). solution

Answer 1

a)
`\vec u_(R) = 0.924\cdot i + 0.355\cdot j + 0.142\cdot k`, b)
`||\vec R|| \approx 32.218`, c)
`\vec R = -580.483\cdot i + 3192.659\cdot j -2902.417\cdot k`

Step-by-step explanation:

a) The resultant vector is obtained by summing all components:

`\vec R = 10\cdot i-4\cdot j + 8\cdot k+16\cdot i + 14\cdot j -4\cdot k`

`\vec R = 26\cdot i + 10\cdot j + 4\cdot k`

Its magnitude is determined herein:

`||\vec R|| = \sqrt{26^(2)+10^(2)+4^(2)}`

`||\vec R|| \approx 28.142`

Unit vector in the given direction is:

`\vec u_(R) = (26)/(28.142)\cdot i + (10)/(28.142)\cdot j + (4)/(28.142)\cdot k`

`\vec u_(R) = 0.924\cdot i + 0.355\cdot j + 0.142\cdot k`

b) The resultant vector is obtained by summing all components:

`\vec R = 5\cdot i + 8\cdot i + 7\cdot j - 2\cdot k+10\cdot i -12\cdot j +24\cdot k`

`\vec R = 23\cdot i - 5\cdot j + 22\cdot k`

Its magnitude is determined herein:

`||\vec R|| = \sqrt{23^(2)+(-5)^(2)+22^(2)}`

`||\vec R|| \approx 32.218`

c) Magnitudes of
`\vec M` and
`\vec N` are, respectively:

`||\vec M|| = \sqrt{(-10)^(2)+4^(2)+(-8)^(2)}`

`||\vec M|| \approx 13.416`

`||\vec N|| = \sqrt{8^(2)+7^(2)+(-2)^(2)}`

`||\vec N|| \approx 10.817`

The sum of both vectors is:

`\vec Q = \vec M + \vec N = -2\cdot i +11\cdot j -10\cdot k`

Finally, the resultant is:

`\vec R = 2||\vec M||\cdot ||\vec N|| \cdot \vec Q`

`\vec R = -580.483\cdot i + 3192.659\cdot j -2902.417\cdot k`