Question

0.8 Find the angle between the vectors A = 5i +6j+4k and B=2i-2j+3k​

Answer 1

Answer:

Explanation:

A.B = ║A║║B║ cos∝

(5i +6j+4k).(2i-2j+3k​) = √77.√17cos∝ = 10 -12 +12 = 10

cos∝ = 10/ (√77.√17)

∝ = 85.62°

Answer 2

The angle between the vectors is 73.97 degrees.

Explanation:

Vector 1, $A=5i+6j+4k$

Vector 2, $B=2i-2j+3k$

To find,

The angle between A and B.

Solution,

Let $\theta$ is the angle between A and B. It can be calculated using concept of dot product as :

$|A|=\sqrt{5^2+6^2+4^2}=8.77$

$|B|=\sqrt{2^2+(-2)^2+3^2}=4.12$

$A{\cdot} B=|A||B|\ cos\theta$

$cos\theta=\dfrac{A{\cdot} B}{|A||B|}$

$cos\theta=\dfrac{(5i+6j+4k){\cdot} (2i-2j+3k)}{8.77\times 4.12}$

$cos\theta=\dfrac{10}{36.13}$

$\theta=73.97^{\circ}$

So, the angle between the vectors is 73.97 degrees.

Learn more

Dot product

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