The angle between two vectors given by 6i¯+6j¯−3k¯ and 7i¯+4j¯+4k¯ is

The angle between two vectors given by $6 \bar{i} + 6 \bar{j} - 3 \bar{k}$ and $7 \bar{i} + 4 \bar{j} + 4 \bar{k}$ is

A
$\cos^{- 1} (\frac{1}{\sqrt{3}})$
B
$\cos^{- 1} (\frac{5}{\sqrt{3}})$
C
$\sin^{- 1} (\frac{2}{\sqrt{3}})$
D
$\sin^{- 1} (\frac{\sqrt{5}}{3})$

Solution:

If the angle between the vectors is ‘ $θ$ ’ then

$\cos θ = \frac{\vec{A} . \vec{B}}{| \vec{A} | . | \vec{B} |}$

$\Rightarrow \cos θ = \frac{42 + 24 - 12}{(\sqrt{36 + 36 + 9}) (\sqrt{49 + 16 + 16})}$

$\cos θ = \frac{54}{(9) (9)} = \frac{2}{3}$

$∴ \sin θ = \frac{\sqrt{5}}{3}$

$\Rightarrow θ = \sin^{- 1} (\frac{\sqrt{5}}{3})$

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