If the vectors b = tanα, - 1, 

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 Multiple Choice QuestionsMultiple Choice Questions

391.

If a = i^ - 2j^5 and b = 2i^ + j^ + 3k^14 are vectors in space, then the value of 2a +b . a × b × a - 2b is

  • 0

  • 1

  • 5

  • 4


392.

The value of i^ . j^ × k^ + j^ . k^ × i^ + k^ . i^ × j^ is

  • 0

  • 1

  • 3

  • - 3


393.

The values of λ, such that (x, y, z) if (0, 0, 0) and i^ + j^ + 3k^x + 3i^ - 3j^ + k^y + - 4i^ + 5j^ are

  • 0, 1

  • - 1, 1

  • - 1, 0

  • - 2, 0


394.

If G and G' are respectively centroid of ABC and A' B' C', then AA' + BB' + CC' is equal to

  • 2GG'

  • 3GG'

  • 23GG'

  • 13GG'


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395.

If a = 3i^ - 4j^ + 5k^, b = i^ + j^ + k^ and c = - 2i^ + 3j^ - 5k^, and if [·] is the least integer function, then [a + b + c] is equal to

  • 1

  • 2

  • 3

  • 0


396.

If a = - i^ + j^ + k^ and b = 2i^ + k^, then the vector satisfyin the following conditions

(i) it is coplanar witha and b,

(ii) it is perpendicular to b and

(iii) a · c = 7, is

  • - i^ + 2j^ + 2k^

  • - 32i^ + 52j^ + 3k^

  • - 3i^ + 5j^ + 6k^

  • - 6i^ + k^


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397.

If the vectors b = tanα, - 1, 2sinα2 and c = tanα, tanα,  - 3sinα2 are orthagonal and  a vector a = 1, 3, sin2α makes an obtuse angle with the Z-axis, then the value of α is

  • 4n + 2π + tan-12

  • 4n + 2π - tan-12

  • 4n + 1π + tan-12

  • 4n + 1π - tan-12


D.

4n + 1π - tan-12

Let b = tanα, - 1, 2sinα2and c = tanα, tanα,  - 3sinα2Since, the vector a = 1, 3, sin2α makes an obtuseangle with the the Z-axis. Therefore, its z-componentis negative.i.e., sin2α < 0        - 1  sin2α  0        ...(i)Since, b and c are orthogonal. b . c = 0 tan2α - tanα - 6 = 0 tanα - 3tanα + 2 = 0 tanα = 3, - 2  tanα = 3Then, sin2α = 2tanα1 + tan2α

= 610 > 0, which is contradiction ton Eq. (i) tanα = 3 is not possible.Thus, tanα = - 2 and for this value of tanα, we ettan2α = 2tanα1 - tan2α = 43Since, sin2α < 0 and tanα > 0Therefore, 2α is in third quadrant.Also, sinα2 is meaningful, if 0 < sinα2 < 1.When these conditions are satisfied, α is given byα = 4n + 1π - tan-12


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398.

r . i^2 + r . j^2 + r . k^2 is equal to

  • 0

  • 1

  • r2

  • 3r2


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399.

The component of i^ + j^ along j^ and k^ will be

  • i^ + j^2

  • j^ + k^2

  • k^ + i^2

  • None of these


400.

If a = 2i^ + 5j^ and b = 2i^ - j^, then the unit vector along a + b will be

  • i^ - j^2

  • i^ + j^

  • 2i^ + j^

  • i^ + j^2


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