Question

Is the resultant of the vectors 3i^ + 4j^ + 5k^ , 5i^ + 3j^ + 4k^ makes an angle with x-axis then find the cos?

Answer 1

A+B=8i +j -k

A-B= -2i +7j -9k

let A+B = C

and A - B = D

then cosine of angle between 2 vectors is given by

cos(theta)=A·B/|A||B|  

A·B= 0  

hence cos(theta)=0

theta = 90°

Answer 2

Given that,

The vector A is

$\vec{A}=3i+4j+5k$

$\vec{B}=5i+3j+4k$

We need to calculate the resultant of the vector

Using formula of resultant vector

$\vec{R}=\vec{A}+\vec{B}$

Put the value into the formula

$\vec{R}=(3i+4j+5k)+(5i+3j+4k)$

$\vec{R}=8i+7j+8k$

The magnitude of the resultant

$R=\sqrt{8^2+7^2+8^2}$

$R=\sqrt{177}$

$R=13.3$

The angle with x-axis.

We need to calculate the value of cosθ

Using formula of angle

$\cos\theta=\dfrac{R\cdot i}{|R|}$

$\cos\theta=\dfrac{(8i+7j+8k)\cdot i}{13.3}$

$\cos\theta=\dfrac{8}{13.3}$

$\cos\theta=0.60$

Hence, The value of cosθ is 0.60.