What is the angle between the following pair vectors- A-k- and B-2-2-2k-

A⃗.B⃗ = AB cos θ =A⃗.B⃗/AB But, A⃗.B⃗= (î+ĵ+k̂).( 2î 2ĵ 2k̂) A⃗.B⃗= 2 2 2= 6 Again A=|A⃗|=√((1)^2+(1)^2+(1)^2) = √(3) B=|B⃗|=√(( 2)^2+( 2)^2 + ( 2)^2)= √(12) ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

What is the a...

Question

What is the angle between the following pair vectors? $\to A =^i +^j +^k a n d \to B = - 2^i - 2^j - 2^k$

$30^{\circ}$

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$60^{\circ}$

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$90^{\circ}$

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$180^{\circ}$

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Solution

The correct option is D
$180^{\circ}$

$\to A . \to B = A B c o s θ = \frac{\to A . \to B}{A B} B u t, \to A . \to B = (^i +^j +^k) . (- 2^i - 2^j - 2^k) \to A . \to B = - 2 - 2 - 2 = - 6 A g a i n A = | \to A | = \sqrt{(1)^{2} + (1)^{2} + (1)^{2}} = \sqrt{3} B = | \to B | = \sqrt{(- 2)^{2} + (- 2)^{2} + (- 2)^{2}} = \sqrt{12} = 2 \sqrt{3} N o w, c o s θ = \frac{- 6}{\sqrt{3} \times 2 \sqrt{3}} = - 1 \Rightarrow θ = 180^{\circ}$

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Q. The angle between the two vectors

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and

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Q. Given

\to a =^i +^j -^k, \to b = -^i + 2^j +^k

and

\to c = -^i + 2^i -^k . A

unit vector perpendicular to both

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\to b + \to c

What is the angle between the following vectors?

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

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

The correct option is D 180∘ →A.→B=AB cos θ=→A.→BABBut, →A.→B=(^i+^j+^k).(−2^i−2^j−2^k)→A.→B=−2−2−2=−6Again A=|→A|=√(1)2+(1)2+(1)2=√3B=|→B|=√(−2)2+(−2)2+(−2)2=√12=2√3Now, cos θ=−6√3×2√3=−1⇒θ=180∘

What is the angle between the following pair vectors? $\to A =^i +^j +^k a n d \to B = - 2^i - 2^j - 2^k$