What is the angle between the following vectors? ⃗ A = ^ i + ^ j + ^ k and ⃗ B = − 2 ^ i − 2 ^ j − 2 ^ k . Options: A . &nbsp 45 ∘ B . &nbsp 90 ∘ C . &nbsp 0 ∘ D . &nbsp 180 ∘

Answer: Option D : D ⃗ A . ⃗ B = | A | | B | c o s θ o r , c o s θ = ⃗ A . ⃗ B | A | | B | ----------------(i) But, ⃗ A . ⃗ B = ( ^ i + ^ j + ^ k ) . ( − 2 ^ i − 2 ^ j − 2 ^ k ) ⃗ A . ⃗ B = − 2 − 2 − 2 ⃗ A . ⃗ B = − 2 − 2 − 2 = − 6 Again A = | ⃗ A | = √ ( 1 ) 2 + ( 1 ) 2 + ( 1 ) 2 = √ 3 ; B = | ⃗ B | = √ ( − 2 ) 2 + ( − 2 ) 2 + ( − 2 ) 2 = √ 12 = 2 √ 3 Now, c o s θ = − 6 √ 3 × 2 √ 3 = − 1 ⇒ θ = 180 ∘

What is the angle between the following vectors? ⃗A=^i+^j+^k and

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Question -

What is the angle between the following vectors?

$⃗ A =^i +^j +^k$ and $⃗ B = - 2^i - 2^j - 2^k$ .

Options:

A .

45^{\circ}

B .

90^{\circ}

C .

0^{\circ}

D .

180^{\circ}

Answer: Option D
:
D

$⃗ A . ⃗ B = | A | | B | c o s θ o r, c o s θ = \frac{⃗ A . ⃗ B}{| A | | B |}$ ----------------(i)

But, $⃗ A . ⃗ B = (^i +^j +^k) . (- 2^i - 2^j - 2^k)$

$⃗ A . ⃗ B = - 2 - 2 - 2$

$⃗ A . ⃗ B = - 2 - 2 - 2 = - 6$

Again A = $| ⃗ A | = \sqrt{(1)^{2} + (1)^{2} + (1)^{2}} = \sqrt{3}$ ;

B = $| ⃗ B | = \sqrt{(- 2)^{2} + (- 2)^{2} + (- 2)^{2}} = \sqrt{12} = 2 \sqrt{3}$

Now, $c o s θ = \frac{- 6}{\sqrt{3} \times 2 \sqrt{3}} = - 1 \Rightarrow θ = 180^{\circ}$

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