Answer to Question #201051 in Mechanics | Relativity for karen

Question

Given the vectors A = 8i + 4j -2k N B = 2j + 6k m C = 3i -2j + 4k m Calculate the following: a. 𝐀 ∙ 𝐁 b. The orthogonal component of B in the direction of C c. The angle between A and C d. A x B e. a unit vector λ that is perpendicular to both A and B f. 𝐀 x 𝐁 ∙ 𝐂

Answer 1

A = 8i + 4j -2k N

B = 2j + 6k m

C = 3i -2j + 4k m

(a) $A.B= (8i + 4j -2k).(2j + 6k)$  

$=(8\times0)+(4\times2)+(-2\times6)$

$=-4$

(b) Orthogonal component of B to C is given by,

$=\dfrac{\vec C\times(\vec B\times\vec C)}{|\vec C|^2}$

$=\dfrac{(3i -2j + 4k)\times((2j + 6k)\times(3i -2j + 4k))}{|3i -2j + 4k|^2}$

$=\dfrac{(3i -2j + 4k)\times(20i+18j-6k)}{29}$

$=\dfrac{-60i+98j+94k}{29}$

(c) Angle between A and C is given by,

$\theta=\cos^{-1}\dfrac{\vec A.\vec C}{|\vec A||\vec C|}$

$\theta=\cos^{-1}\dfrac{(8i + 4j -2k).(3i -2j + 4k)}{|8i + 4j -2k||3i -2j + 4k|}$

$=\cos^{-1}\dfrac{8}{\sqrt{84\times29}}$

$\theta=80.67\degree$

(d) $\vec A\times \vec B=(8i + 4j -2k)\times( 2j + 6k)$

$=28i-48j+16k$

(e) Unit vector perpendicular to both A and B is given by,

$\lambda=\dfrac{\vec A\times\vec B}{|\vec A\times\vec B|}$

$\lambda=\dfrac{28i-48j+16k}{|28i-48j+16k|}$

$\lambda=\dfrac{28i-48j+16k}{3344}$

(f) $(\vec A\times\vec B).\vec C=(28i-48j+16k).(3i -2j + 4k)$

$=244$