What is the angle between the following vectors? → A = 3 ^ i − 2 ^ j + ^ k → B = 2 ^ i + 6 ^ j − 6 ^ k Options: A . &nbsp c o s − 1 ( 12 √ 266 ) B . &nbsp c o s − 1 ( − 6 √ 266 ) C . &nbsp c o s − 1 ( − 6 √ 1064 ) D . &nbsp None of these

Answer: Option B : B → A . → B = ( 3 × 2 ) + ( − 2 × 6 ) + ( 1 × − 6 ) = 6 - 12 - 6 = -12 | → A | = √ 3 2 + ( − 2 ) 2 + 1 2 = √ 9 + 4 + 1 = √ 14 | → B | = √ 2 2 + 6 2 + ( − 6 ) 2 = √ 4 + 36 + 36 = √ 76 → A . → B | → A | | → B | = − 12 √ 14 √ 76 = − 6 √ 266 ∴ c o s θ = − 6 √ 266 ⇒ θ = c o s − 1 ( − 6 √ 266 )

What is the angle between the following vectors? →A=3^i−2^j+^k

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Question -

What is the angle between the following vectors?

$\to A = 3^i - 2^j +^k$

$\to B = 2^i + 6^j - 6^k$

Options:

A .

c o s^{- 1} (\frac{12}{\sqrt{266}})

B .

c o s^{- 1} (\frac{- 6}{\sqrt{266}})

C .

c o s^{- 1} (\frac{- 6}{\sqrt{1064}})

D . None of these

Answer: Option B
:
B

$\to A . \to B = (3 \times 2) + (- 2 \times 6) + (1 \times - 6)$

= 6 - 12 - 6

= -12

$| \to A | = \sqrt{3^{2} + (- 2)^{2} + 1^{2}} = \sqrt{9 + 4 + 1} = \sqrt{14}$

$| \to B | = \sqrt{2^{2} + 6^{2} + (- 6)^{2}} = \sqrt{4 + 36 + 36} = \sqrt{76}$

$\frac{\to A . \to B}{| \to A | | \to B |} = \frac{- 12}{\sqrt{14} \sqrt{76}} = \frac{- 6}{\sqrt{266}}$

$∴ c o s θ = \frac{- 6}{\sqrt{266}}$

$\Rightarrow θ = c o s^{- 1} (\frac{- 6}{\sqrt{266}})$

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