IIT - JEE Main Important Questions of Vector Algebra | Zigya

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 Multiple Choice QuestionsMultiple Choice Questions

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1.

The Boolean Expression (p∧~q)∨q∨(~p∧q) is equivalent to:

  • ~p ∧ q

  • p ∧ q

  • p ∨ q

  • p ∨ q

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2.

Let, a, b and c be three non-zero vectors such that no two of them are collinear and left parenthesis straight a space xb right parenthesis space straight x space straight c space equals space 1 third vertical line straight b vertical line vertical line straight c vertical line straight a. if θ is the angle between vectors b and  c, then a value of sin θ is

  • fraction numerator 2 square root of 2 over denominator 3 end fraction
  • fraction numerator negative square root of 2 over denominator 3 end fraction
  • 2 over 3
  • 2 over 3
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3.

If the vectors AB = 3î + 4k̂ and AC = 5î - 2ĵ + 4k̂ are the sides of a Δ ABC, then the length of the median through A is 

  • √18

  • √72

  • √33

  • √45

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4.

Let straight a with hat on top space and space straight b with hat on top be  two unit vectors. If the vectors straight c equals space straight a with hat on top space plus 2 straight b with hat on top and straight d space equals space 5 straight a with hat on top space minus 4 straight b with hat on top are perpendicular to each other, then the angle between straight a with hat on top space and space straight b with hat on top is 

  • π/6

  • π/2

  • π/3

  • π/3

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5.

Let ABCD be a parallelogram such that AB with rightwards arrow on top space equals space straight q with rightwards arrow on top comma space AD with rightwards arrow on top space equals space straight p with rightwards arrow on top and ∠BAD be an acute angle. If straight r with rightwards arrow on top  is the vector that coincides with the altitude directed from the vertex B the side AD, then straight r with rightwards arrow on top is given byLet ABCD be a parallelogram such that AB = q,AD = p and ∠BAD be an acute angle. If r is the vector that coincides with the altitude directed from the vertex B to the side AD, then r is given by (1)

  • straight r with rightwards arrow on top space equals space 3 straight q with rightwards arrow on top space minus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
  • straight r with rightwards arrow on top space equals negative space straight q with rightwards arrow on top space plus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
  • straight r with rightwards arrow on top space equals space straight q with rightwards arrow on top space minus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
  • straight r with rightwards arrow on top space equals space straight q with rightwards arrow on top space minus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
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6. If space straight a with rightwards arrow on top space equals space fraction numerator 1 over denominator square root of 10 end fraction space left parenthesis 3 straight i with hat on top space plus straight k with hat on top right parenthesis space and space straight b with rightwards arrow on top space equals space 1 over 7 left parenthesis 2 straight i with hat on top space plus space 3 straight j with hat on top space minus 6 straight k with hat on top right parenthesis comma space then space the space value space of space left parenthesis 2 straight a with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis. left square bracket left parenthesis straight a with rightwards arrow on top space straight x space straight b with rightwards arrow on top right parenthesis space straight x left parenthesis straight a with rightwards arrow on top plus 2 straight b with rightwards arrow on top right parenthesis right square bracket space is
  • -5

  • -3

  • 5

  • 5

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7.

The vector straight a with rightwards arrow on top space and space straight b with rightwards arrow on top are not perpendicular  and straight c with rightwards arrow on top space and space straight d with rightwards arrow on top are two vectors satisfying: straight b with rightwards arrow on top space straight x straight c with rightwards arrow on top space equals space straight b with rightwards arrow on top space straight x straight d with rightwards arrow on top space and space straight a with rightwards arrow on top. straight b with rightwards arrow on top space equals space 0. The vector straight d with rightwards arrow on top is equal to

  • straight b with rightwards arrow on top space minus space open parentheses fraction numerator straight b. straight c over denominator straight a with rightwards arrow on top. straight d with rightwards arrow on top end fraction close parentheses straight c with rightwards arrow on top
  • straight c with rightwards arrow on top space plus open parentheses fraction numerator straight a with rightwards arrow on top. straight c with rightwards arrow on top over denominator straight a with rightwards arrow on top. straight b with rightwards arrow on top end fraction close parentheses straight b with rightwards harpoon with barb upwards on top
  • straight b with rightwards arrow on top space plus space open parentheses fraction numerator straight b with rightwards arrow on top. straight c with rightwards arrow on top over denominator straight a with rightwards arrow on top. straight b with rightwards arrow on top end fraction close parentheses straight c with rightwards arrow on top
  • straight b with rightwards arrow on top space plus space open parentheses fraction numerator straight b with rightwards arrow on top. straight c with rightwards arrow on top over denominator straight a with rightwards arrow on top. straight b with rightwards arrow on top end fraction close parentheses straight c with rightwards arrow on top
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8.

The circle x2+ y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if

  • -85 < m < -35

  • -35 < m < 15

  • 15 < m < 65

  • 15 < m < 65

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9.

Let bold a bold space bold equals bold j with bold hat on top bold minus bold k with bold hat on top bold space and bold c bold space bold equals bold space bold i with bold hat on top bold minus bold j with bold hat on top bold minus bold k with bold hat on top. Then, the vector b satisfying a x b + c = 0  and a.b  = 3

  • negative straight i with hat on top plus straight j with hat on top space minus 2 straight k with hat on top
  • 2 straight i with hat on top minus straight j with hat on top space plus 2 straight k with hat on top
  • negative straight i with hat on top minus straight j with hat on top space minus 2 straight k with hat on top
  • negative straight i with hat on top minus straight j with hat on top space minus 2 straight k with hat on top
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10.

If the vectors bold a bold space bold equals bold space bold i with bold hat on top bold space bold minus bold j with bold hat on top bold space bold plus bold 2 bold k with bold hat on top bold comma bold space bold b bold space bold equals bold 2 bold i with bold hat on top bold space bold plus bold 4 bold j with bold hat on top bold space bold plus bold k with bold hat on top bold space bold and bold space bold c bold space bold equals bold space bold lambda bold i with bold hat on top bold space bold plus bold j with bold hat on top bold space bold plus bold mu bold k with bold hat on top are mutually orthogonal, then (λ,μ) is equal to

  • (-3,2)

  • (2,-3)

  • (-2,3)

  • (-2,3)

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