Work Energy And Power MCQs for NEET — Physics Questions with Answers

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A→=i^+j^-k^ ; B→=2i^+3j^+5k^ angle between A→ and B→ is

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Explanation

Angle between two vectors = cos-1A→.B→AB

=cos-12+3-53×38=cos-1(0)=90°

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Given the vectors

                         A→=2i^+3j^-k^                         B→=3i^-2j^-2k^&                      C→=pi^+pj^+2pk^

Find the angle between A→-B→ & C→

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Explanation

(C)

 

A→-B→= 2i^+3j^-k^-3i^-2j^-2k^          = -i^+5j^+k^Let A→-B→= D→   let∴ D→= -i^+5j^+k^     D→= -12+52+12    = 27To find the angle between D→ & C→C→= p2+p2+2p2   = P6D→. C→= DC cosθ-i^+5j^+k^. pi^+pj^+2pk^= 27×p6 cosθ-p+5p+2p= p27×6 cosθ6= 27×6 cosθcosθ= 627         = 29cosθ= 23    or    θ=cos-123

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The vector having a magnitude of 10 and perpendicular to the vector 3i^-4j^ is 

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Explanation

3i+4j.8i+6j=0

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A force F→=3i^+cj^+2k^ acting on a particle causes a displacement d→=-4i^+2j^+3k^. If the work done is 6J then the value of 'c' is

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Explanation

W=F→.d→⇒6=F→.d→⇒6=-12+2C+6⇒C=6

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A vector that is perpendicular to both the vectors a→=i^-2j^+k^ and b→=i^-j^+k^ is

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Explanation

(A)

 

 

i^+k^ = -i^+0j^+k^a→.-i^+k^ = -1+0+1= 0b→.-i^+k^ = -1+0+1= 0

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If θ is the angle between vectors a→ and b→, and |a→×b→|=a→.b→, then θ is equal to

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Explanation

Given, |a→×b→|=a→.b→⇒ab sinθ=ab cosθ⇒tan θ=1⇒θ=45°

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The vector b→, which is collinear with the vector a→=(2, 1, -1) and satisfies the condition a→.b→=3, is 

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Explanation

Going through the options;Only (A) satifies the condition a→.b→=3.

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If 3i^+2j^+8k^ and 2i^+xj^+k^ are at right angle then x=

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Explanation

If two vectors are at right angle, then their dot product = 0

⇒(3i^+2j^+8k^).(2i^+xj^+k^)=0⇒6+2x+8=0⇒x=-7

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If a, b, c are three non-zero vectors such that a→+b→+c→=0 the value of a→.b→+b→.c→+c→.a→ is

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Let α, β, γ be distinct real numbers. The points with position vectors αi^+βj^+γk^ , βi^+γj^+αk^ , γi^+αj^+βk^

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