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

Respuesta :

Answer:

a) [tex]\vec u_{R} = 0.924\cdot i + 0.355\cdot j + 0.142\cdot k[/tex], b) [tex]||\vec R|| \approx 32.218[/tex], c) [tex]\vec R = -580.483\cdot i + 3192.659\cdot j -2902.417\cdot k[/tex]

Explanation:

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

[tex]\vec R = 10\cdot i-4\cdot j + 8\cdot k+16\cdot i + 14\cdot j -4\cdot k[/tex]

[tex]\vec R = 26\cdot i + 10\cdot j + 4\cdot k[/tex]

Its magnitude is determined herein:

[tex]||\vec R|| = \sqrt{26^{2}+10^{2}+4^{2}}[/tex]

[tex]||\vec R|| \approx 28.142[/tex]

Unit vector in the given direction is:

[tex]\vec u_{R} = \frac{26}{28.142}\cdot i + \frac{10}{28.142}\cdot j + \frac{4}{28.142}\cdot k[/tex]

[tex]\vec u_{R} = 0.924\cdot i + 0.355\cdot j + 0.142\cdot k[/tex]

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

[tex]\vec R = 5\cdot i + 8\cdot i + 7\cdot j - 2\cdot k+10\cdot i -12\cdot j +24\cdot k[/tex]

[tex]\vec R = 23\cdot i - 5\cdot j + 22\cdot k[/tex]

Its magnitude is determined herein:

[tex]||\vec R|| = \sqrt{23^{2}+(-5)^{2}+22^{2}}[/tex]

[tex]||\vec R|| \approx 32.218[/tex]

c) Magnitudes of [tex]\vec M[/tex] and [tex]\vec N[/tex] are, respectively:

[tex]||\vec M|| = \sqrt{(-10)^{2}+4^{2}+(-8)^{2}}[/tex]

[tex]||\vec M|| \approx 13.416[/tex]

[tex]||\vec N|| = \sqrt{8^{2}+7^{2}+(-2)^{2}}[/tex]

[tex]||\vec N|| \approx 10.817[/tex]

The sum of both vectors is:

[tex]\vec Q = \vec M + \vec N = -2\cdot i +11\cdot j -10\cdot k[/tex]

Finally, the resultant is:

[tex]\vec R = 2||\vec M||\cdot ||\vec N|| \cdot \vec Q[/tex]

[tex]\vec R = -580.483\cdot i + 3192.659\cdot j -2902.417\cdot k[/tex]