Answer:
|a-b|=6.34m
Explanation:
To find the difference between the vector you can use the formulas for the magnitude of a vector and also for the doc product between two vectors:
[tex]a=\sqrt{a_1^2+a_2^2+a_3^2}\\\\b=\sqrt{b_1^2+b_2^2+b_3^2}\\\\\vec{a}\cdot\vec{b}=abcos\theta\\\\a_1b_1+a_2b_2+a_3b_3=abcos\theta[/tex]
a1, a2, a3: components of a vector
b1, b2, b3: components of b vector
a: magnitude of a = 2m
b: magnitude of b = 8m
angle = 30°
By squaring the first two equations
[tex]a^2=4m^2=a_1^2+a_2^2+a_3^2\\\\b^2=64m^2=b_1^2+b_2^2+b_3^2\\\\[/tex]
Then, you multiply by 2 the third equation:
[tex]2a_1b_1+2a_2b_2+2a_3b_3=2(2m)(8m)cos30\°=27.71m^2[/tex]
Now, you sum the first two equations ans take the difference with the third equation. Thus, you obtain a perfect square trinomial:
[tex](a_1^2-2a_1b_1+b_1^2)+(a_2^2-2a_2b_2+b_2^2)+(a_3^2-2a_3b_3+b_3^2)=(4+64-27.71)m^2\\\\(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2=40.29m^2[/tex]
This last expression is the square of the magnitude of the difference a-b. Hence you have:
[tex]\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_2-b_2)^2}=\sqrt{40.29m^2}=6.34m[/tex]
thus, the magnitude of the difference is 6.34m
