If the position vectors of the vertices of atriangle are 2i^ 

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 Multiple Choice QuestionsMultiple Choice Questions

401.

For any vectors, a, b, c a × b + c + b × c + a + c × a +b = ?

  • 0

  • a + b + c

  • [a, b, c]

  • a × b × c


402.

If a = i^ + 4j^, b = 2i^ - 2j^, c = 5i^ + 9j^, then c is equal to

  • 2a + b

  • a + 2b

  • 3a + b

  • a + 3b


403.

If a = i^ + j^ + tk^, b = i^ + 2j^ + 3k^ then the values of 't' for which (a + b) and (a - b) are perpendicular, are

  • ± 2

  • ± 23

  • ±32

  • ± 3


404.

i^ - j^, j^ - k^, k^ - i^ is equal to

  • 0

  • 1

  • 2

  • 3


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405.

If a . i^ = a . i^ + j^ = ai^ + j^ + k^, then a is equal to

  • i^

  • j^

  • k^

  • i^ + j^ + k^


406.

If three points A,B,and C have position vectors 1, x, 3, 3, 4, 7 and y, - 2, - 5 respectively and if they are collinear, then (x, y) is

  • (2, - 3)

  • ( - 2, 3)

  • ( - 2, - 3)

  • (2, 3)


407.

The orthogonal projection of a on b is

  • a . bba2

  • a . bbb2

  • aa2

  • bb


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408.

If the position vectors of the vertices of atriangle are 2i^ - j^ + k^i^ - 3j^ - 5k^ and 3i^ - 4j^ - 4k^, then the probability of triangle is

  • equilateral

  • isosceless

  • right angled isosceless

  • right angled


D.

right angled

Let A = 2i^ - j^ + k^, B = i^ - 3j^ - 5k^C = 3i^ - 4j^ - 4k^AB = - i^ - 2j^ - 6k^, BC = 2i^ - j^ + k^and CA = - i^ + 3j^ + 5k^a = BC = 4 + 1 + 1 = 6b = CA = 1 + 9 + 25 = 35c = AB = 1 + 4 + 36 = 41Now,    c2 = a2 + b2 41 = 6 + 35     41 = 41 It is right angled triangle.


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409.

If [a b c] = 3, then the volume (in cubic units) of e parallelepiped with 2a + b, 2b + c and 2c + a as edges, is

  • 15

  • 22

  • 25

  • 27


410.

a +b . b +c × a + b +c is equal to

  • 0

  • - [a b c]

  • 2[a b c]

  • [a b c]


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