If with reference to a right handed system of mutually perpendic

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 Multiple Choice QuestionsLong Answer Type

291.

If straight a with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top space and straight b with rightwards arrow on top space equals space straight j with hat on top space minus space straight k with hat on top.  Find a vector straight c with rightwards arrow on top such that straight a with rightwards arrow on top cross times space straight c with rightwards arrow on top space equals space straight b with rightwards arrow on top and straight a with rightwards arrow on top. space straight c with rightwards arrow on top space equals space 3.

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292.

Let straight a with rightwards arrow on top space equals straight i with hat on top space plus space 4 space straight j with hat on top space plus space 2 space straight k with hat on top comma space space space space space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space space 7 space straight k with hat on top and straight c with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space 4 space straight k with hat on top. Find a vector straight d with rightwards arrow on top which is perpendicular to both straight a with rightwards arrow on top space and space straight b with rightwards arrow on top comma space space straight c with rightwards arrow on top. space straight d with rightwards arrow on top space equals space 15.

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293. If with reference to a right handed system of mutually perpendicular unit vector unit vector straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top

express straight beta with rightwards arrow on top in the form straight beta with rightwards arrow on top in the form straight beta with rightwards arrow on top space equals space straight beta with rightwards arrow on top subscript 1 space equals space straight beta with rightwards arrow on top subscript 1 space plus space straight beta with rightwards arrow on top subscript 2 space end subscript space where space stack straight beta subscript 1 with rightwards arrow on top space is space parallel space to space straight alpha space and space stack straight beta subscript 2 with rightwards arrow on top space is space perpendicular space to space straight alpha with rightwards arrow on top.


Let straight alpha with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space straight j with hat on top comma space space space space straight beta with rightwards arrow on top space equals space 2 straight i with hat on top plus straight j with hat on top space minus space 3 straight k with hat on top
Let  stack straight beta subscript 1 with rightwards arrow on top space equals space straight a subscript 1 straight i with hat on top space plus space straight a subscript 2 straight j with hat on top space plus space straight a subscript 3 straight k with hat on top comma space space stack straight beta subscript 2 with rightwards arrow on top space equals space straight b subscript 1 straight i with hat on top space plus space straight b subscript 2 straight j with hat on top space plus space straight b subscript 3 straight k with hat on top
therefore space space space space space straight beta with rightwards arrow on top space equals straight beta with rightwards arrow on top subscript 1 plus space stack straight beta subscript 2 with rightwards arrow on top space equals space left parenthesis straight a subscript 1 plus straight b subscript 1 right parenthesis space straight i with hat on top space plus space left parenthesis straight a subscript 2 plus straight b subscript 2 right parenthesis straight j with hat on top space plus space left parenthesis straight a subscript 3 plus straight b subscript 3 right parenthesis straight k with hat on top
rightwards double arrow space space 2 straight i with hat on top space plus space straight j with hat on top space minus space 3 space straight k with hat on top space equals space left parenthesis straight a subscript 1 plus straight b subscript 1 right parenthesis straight i with hat on top space plus space left parenthesis straight a subscript 2 plus straight b subscript 2 right parenthesis space straight j with hat on top space plus space left parenthesis straight a subscript 3 plus straight b subscript 3 right parenthesis space straight k with hat on top
rightwards double arrow space space space straight a subscript 1 plus straight b subscript 1 space equals space 2 comma space space space space space straight a subscript 2 plus straight b subscript 2 space equals space 1 comma space space straight a subscript 3 plus straight b subscript 3 space equals space minus 3
therefore space space space space space space straight b subscript 1 space equals space 2 space minus straight a subscript 1 comma space space space space space straight b subscript 2 space equals space 1 minus straight a subscript 2 comma space space space straight b subscript 3 space equals space minus 3 minus straight a subscript 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space space space space space space space end subscript
Since stack straight beta subscript 1 with rightwards arrow on top is parallel to straight alpha with rightwards arrow on top
 therefore space space space space space space straight beta with rightwards arrow on top subscript 1 space cross times space straight alpha with rightwards arrow on top space equals space 0 space space space space space space rightwards double arrow space space space space space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row cell straight a subscript 1 end cell cell straight a subscript 2 end cell cell straight a subscript 3 end cell row 3 cell negative 1 end cell 0 end table close vertical bar space equals space 0 with rightwards arrow on top
rightwards double arrow space space space space space space straight a subscript 3 straight i with hat on top space plus space 3 space straight a subscript 3 straight j with hat on top space plus left parenthesis negative straight a subscript 1 space minus space 3 space straight a subscript 2 right parenthesis space straight k with hat on top space equals space 0 with rightwards arrow on top
rightwards double arrow space space space space space space straight a subscript 3 space equals space 0 comma space space space space 3 space straight a subscript 3 space equals space 0 comma space space space space space minus straight a subscript 1 space minus space 3 space straight a subscript 2 space equals space 0
rightwards double arrow space space space space space straight a subscript 3 space equals space 0 space space space straight a subscript 1 space equals space minus 3 space straight a subscript 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
Again comma space space stack straight beta subscript 2 with rightwards arrow on top space is space perpendicular space to space straight alpha with rightwards arrow on top
therefore space space space space space stack straight beta subscript 2 with rightwards arrow on top space. space straight alpha with rightwards arrow on top space equals space 0 space space space space space space rightwards double arrow space space space space left parenthesis straight b subscript 1 straight i with hat on top space plus space straight b subscript 2 straight j with hat on top space plus space straight b subscript 3 straight k with hat on top right parenthesis. space space space left parenthesis 3 space straight i with hat on top space minus space straight j with hat on top right parenthesis space equals space 0
rightwards double arrow space space space space 3 space straight b subscript 1 space minus space straight b subscript 2 space equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
rightwards double arrow space 3 space left parenthesis 2 minus straight a subscript 1 right parenthesis space minus space 1 left parenthesis 1 minus straight a subscript 2 right parenthesis space equals space 0 space space space space space space rightwards double arrow space space space space space 6 minus 3 space straight a subscript 1 space minus space 1 space plus space straight a subscript 2 space equals space 0
rightwards double arrow space space space 5 minus 3 space straight a subscript 1 space plus space straight a subscript 2 space equals space 0 space space space space space space
rightwards double arrow space space space 5 plus 9 space straight a subscript 2 plus space straight a subscript 2 space space space equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 2 right parenthesis close square brackets
rightwards double arrow space space space space space 10 space straight a subscript 2 space equals space minus 5 space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space space space straight a subscript 2 space equals space minus 1 half space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
therefore space space space space space straight a subscript 1 space equals 3 over 2                                                                               open square brackets because space of space left parenthesis 1 right parenthesis close square brackets

therefore space space space space straight beta with rightwards arrow on top subscript 1 space equals space 3 over 2 straight i with hat on top space minus space 1 half straight j with hat on top
From (1),   straight b subscript 1 space equals 2 minus 3 over 2 space equals space 1 half comma space space space space space space space straight b subscript 2 equals space 1 plus 1 half space equals 3 over 2 comma space space straight b subscript 3 space equals space minus 3 minus 0 space equals space minus 3

therefore space space stack straight beta subscript 2 with rightwards arrow on top space equals space 1 half straight i with hat on top space plus space 3 over 2 straight j with hat on top space minus space 3 space straight k with hat on top.
 



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 Multiple Choice QuestionsShort Answer Type

294. straight A with rightwards arrow on top comma space straight B with rightwards arrow on top comma space straight C with rightwards arrow on top are unit vectors,  Suppose that  straight A with rightwards arrow on top. space straight B with rightwards arrow on top space equals space straight A with rightwards arrow on top. space straight C with rightwards arrow on top space equals space 0 and the angle between B and C is straight pi over 6. Prove that straight A with rightwards arrow on top space equals plus-or-minus 2 space open parentheses straight B with rightwards arrow on top space cross times space straight C with rightwards arrow on top close parentheses.
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 Multiple Choice QuestionsLong Answer Type

295. Give space space space straight a with rightwards arrow on top space equals space 1 over 7 left parenthesis 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top right parenthesis comma space space straight b with rightwards arrow on top space equals space 1 over 7 left parenthesis 3 space straight i with hat on top space minus space 6 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis comma
space space space space space space space space space space space space space space straight c with rightwards arrow on top space equals space 1 over 7 left parenthesis 6 space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top right parenthesis comma
straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top being a right-handed orthogonal system of unit vectors in space. Show that straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top in another such system.
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 Multiple Choice QuestionsShort Answer Type

296. Find the value of the following:
straight i with hat on top. space open parentheses straight j with hat on top space cross times space straight k with hat on top close parentheses space plus space straight j with hat on top. space space left parenthesis straight i with hat on top space cross times space straight k with hat on top right parenthesis space plus space space straight k with hat on top. space space left parenthesis straight i with hat on top space cross times space straight j with hat on top right parenthesis
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297. Find the value of the following:
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 cross times space left parenthesis straight k with hat on top space cross times space straight i with hat on top right parenthesis space plus space straight k with hat on top space cross times space left parenthesis straight i with hat on top space cross times space straight j with hat on top right parenthesis
space
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298. Find the value of the following:
straight i with hat on top space cross times 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 cross times space left parenthesis straight k with hat on top space cross times space straight i with hat on top right parenthesis space plus space straight k with hat on top space cross times space left parenthesis straight i with hat on top space cross times space straight j with hat on top right parenthesis
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 Multiple Choice QuestionsMultiple Choice Questions

299.

Let vectors straight a with rightwards arrow on top space and space straight b with rightwards arrow on top be such that open vertical bar straight a with rightwards arrow on top close vertical bar space equals 3 space space space and space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space fraction numerator square root of 2 over denominator 3 end fraction comma then straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top is a unit vector, if the angle betweeen straight a with rightwards arrow on top space and space straight b with rightwards arrow on top is 

  • straight pi divided by 6
  • straight pi divided by 4
  • straight pi divided by 3
  • straight pi divided by 3
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300.

If straight theta is the angle between any two vectors straight a with rightwards arrow on top space and space straight b with rightwards arrow on top comma then open vertical bar straight a with rightwards arrow on top. space straight b with rightwards arrow on top close vertical bar space equals space open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar when straight theta is equal to 

  • 0

  • straight pi over 4
  • straight pi over 2
  • straight pi over 2
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