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 Multiple Choice QuestionsMultiple Choice Questions

301.

The value of straight i with hat on top. space left parenthesis straight j with hat on top space cross times space straight k with hat on top right parenthesis. space plus space straight j with hat on top. space left parenthesis straight i with hat on top space cross times space straight k with hat on top right parenthesis space plus space straight k with hat on top. space left parenthesis straight i with hat on top space cross times space straight j with hat on top right parenthesis is

  • 0

  • -1

  • 1

  • 1

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302. Area of a rectangle having vertices A, B, C and D with position vectors
negative straight i with hat on top space plus space 1 half straight j with hat on top space plus space 4 space straight k with hat on top comma space space space straight i with hat on top space plus space 1 half straight j with hat on top space plus space 4 space straight k with hat on top comma space space space space straight i with hat on top space minus space 1 half straight j with hat on top space plus 4 straight k with hat on top space and space minus straight i with hat on top space minus space 1 half straight j with hat on top space plus space 4 straight k with hat on top comma respectively is
  • 1 half
  • 1

  • 2

  • 2

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 Multiple Choice QuestionsLong Answer Type

303. If A, B and C are the vertices of a triangle ABC, prove Sine Formula that
fraction numerator straight a over denominator sin space straight A end fraction space equals space fraction numerator straight b over denominator sin space straight B end fraction space equals space fraction numerator straight c over denominator sin space straight C end fraction.
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304. Using vectors, prove that sin (A + B) = sin A + cos B + cos A sin B. 

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305. Prove by vector method that sin (A – B) = sin A cos B – cos A sin B
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306. Prove that the parallelograms on the same base and between the same parallels are equal in area.
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307. If D, E, F are the mid points of the sides of triangle ABC, prove that
area left parenthesis increment DEF right parenthesis space equals space 1 fourth area space left parenthesis increment thin space ABC right parenthesis.
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 Multiple Choice QuestionsShort Answer Type

308. find space straight lambda space and space straight mu space if

open parentheses straight i with hat on top plus space 3 space straight j with hat on top space plus space 9 space straight k with hat on top close parentheses space space straight x space space left parenthesis 3 space straight i with hat on top plus straight lambda space straight j with hat on top plus space straight mu space straight k with hat on top right parenthesis space equals space 0
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309. If space straight a with rightwards arrow on top equals 4 space straight i with hat on top space minus straight j with hat on top plus space straight k with hat on top space and space straight b with rightwards arrow on top space equals space 2 space straight i with hat on top space minus 2 straight j with hat on top plus space straight k with hat on top comma then space find space straight a space vector space parallel space to space the space vector space straight a with rightwards arrow on top plus straight b with rightwards arrow on top.


Given comma
straight a with rightwards arrow space on top equals space 4 space straight i with hat on top space minus straight j with hat on top plus space straight k with hat on top space and space straight b with rightwards arrow on top space equals space 2 space straight i with hat on top space minus 2 straight j with hat on top plus space straight k with hat on top

therefore space straight a with rightwards arrow on top plus straight b with rightwards arrow on top space equals space open parentheses 4 space straight i with hat on top space minus straight j with hat on top plus space straight k with hat on top close parentheses space plus space open parentheses 2 space straight i with hat on top space minus 2 straight j with hat on top plus space straight k with hat on top close parentheses

equals 6 space straight i with hat on top space minus space 3 straight j with hat on top space plus space 2 straight k with hat on top
unit space vector space parallel space to space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space space equals space fraction numerator straight a with rightwards arrow on top plus straight b with rightwards arrow on top over denominator vertical line straight a with rightwards arrow on top plus straight b with rightwards arrow on top vertical line end fraction
equals space fraction numerator 6 space straight i with hat on top minus 3 straight j with hat on top plus 2 straight k with hat on top over denominator square root of 36 plus 9 plus 4 end root end fraction

equals space fraction numerator 6 space straight i with hat on top minus 3 straight j with hat on top plus 2 straight k with hat on top over denominator square root of 49 end fraction

equals fraction numerator 6 space straight i with hat on top over denominator 7 end fraction minus fraction numerator 3 straight j with hat on top over denominator 7 end fraction plus fraction numerator 2 straight k with hat on top over denominator 7 end fraction
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310.

If straight a with rightwards arrow on top space equals space 2 straight i with hat on top plus straight j with hat on top plus 3 straight k with hat on top space space space and space straight b with rightwards arrow on top space equals space 3 straight i with hat on top space plus space 5 straight j with hat on top space minus space 2 straight k with hat on top comma space then space find space open vertical bar straight a with rightwards arrow on top cross times space straight b with rightwards arrow on top close vertical bar.

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