Question

Cual es el modulo del vector diferencia (a - b) entre dos vectores que forman un angulo de 30° entre si y cuyos módulos son 2m y 8m. Grafica (Les agradecería muchísimo si me ayudan)

Answer 1

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:

$a=\sqrt{a_1^2+a_2^2+a_3^2}\\\\b=\sqrt{b_1^2+b_2^2+b_3^2}\\\\\vec{a}\cdot\vec{b}=abcos\theta\\\\a_1b_1+a_2b_2+a_3b_3=abcos\theta$

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

$a^2=4m^2=a_1^2+a_2^2+a_3^2\\\\b^2=64m^2=b_1^2+b_2^2+b_3^2$

Then, you multiply by 2 the third equation:

$2a_1b_1+2a_2b_2+2a_3b_3=2(2m)(8m)cos30\°=27.71m^2$

Now, you sum the first two equations ans take the difference with the third equation. Thus, you obtain a perfect square trinomial:

$(a_1^2-2a_1b_1+b_1^2)+(a_2^2-2a_2b_2+b_2^2)+(a_3^2-2a_3b_3+b_3^2)=(4+64-27.71)m^2\\\\(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2=40.29m^2$

This last expression is the square of the magnitude of the difference a-b. Hence you have:

$\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_2-b_2)^2}=\sqrt{40.29m^2}=6.34m$

thus, the magnitude of the difference is 6.34m