from Mathematics Vector Algebra Class 12 Maharashtra Board
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Find the value of 'p' for which the vectors 3 straight i with hat on top plus 2 straight j with hat on top plus 9 straight k with hat on top space and space straight i with hat on top minus 2 straight p straight j with hat on top plus 3 straight k with hat on top are parallel.


Since the vectors are parallel, we have
straight a with rightwards harpoon with barb upwards on top equals straight lambda straight b with rightwards harpoon with barb upwards on top
rightwards double arrow 3 straight i with hat on top plus 2 straight j with hat on top plus 9 straight k space equals space straight lambda open parentheses straight i minus 2 straight p straight j with hat on top space plus 3 straight k close parentheses
rightwards double arrow 3 straight i with hat on top plus 2 straight j with hat on top plus 9 straight k space equals space straight lambda straight i with hat on top minus 2 λp straight j with hat on top plus 3 λk
Comparing space the space respective space coefficients comma space we space have
rightwards double arrow space space straight lambda space equals space 3 semicolon
minus 2 λp space equals space 2
rightwards double arrow negative 2 cross times 3 cross times straight p space equals space 2
rightwards double arrow straight p equals fraction numerator negative 1 over denominator 3 end fraction

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Find the angle between the vector straight i with hat on top minus straight j with hat on top space and space straight j with hat on top space minus space straight k with hat on top.


 Let space straight a with rightwards arrow on top space equals space straight i with hat on top minus stack straight j semicolon with hat on top space space 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
straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space left parenthesis straight i with hat on top minus straight j with hat on top right parenthesis space left parenthesis straight j with hat on top space minus space straight k with hat on top right parenthesis space equals space 1 cross times 0 space plus left parenthesis negative 1 right parenthesis cross times 1 space plus 0 space cross times left parenthesis negative 1 right parenthesis space equals space minus 1
open vertical bar straight a with rightwards arrow on top close vertical bar space equals space square root of 1 squared plus left parenthesis negative 1 right parenthesis squared plus 0 squared end root space equals square root of 2
open vertical bar straight b with rightwards arrow on top close vertical bar space equals square root of 0 squared plus 1 squared plus left parenthesis negative 1 right parenthesis squared end root space equals square root of 2
We space know space that space straight a with rightwards arrow on top. straight b with rightwards arrow on top space equals open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar cos space straight theta
Thus comma space cosθ space equals space fraction numerator straight a with rightwards arrow on top. straight b with rightwards arrow on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar. open vertical bar straight b with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator negative 1 over denominator square root of 2 cross times square root of 2 end fraction equals fraction numerator negative 1 over denominator 2 end fraction
rightwards double arrow space space space cosθ space equals space cos 120 degree
rightwards double arrow space space space space straight theta space equals space 120 degree

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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.


Given space that space straight a with rightwards arrow on top space equals space 2 straight i with hat on top space plus space straight j with hat on top space plus space 3 straight k with hat on top 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
We space need space to space find 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
straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals 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 2 1 3 row 3 5 cell negative 2 end cell end table close vertical bar
space space space equals space straight i with hat on top left parenthesis negative 2 minus 15 right parenthesis space minus space straight j with hat on top left parenthesis negative 4 minus 9 right parenthesis space plus space straight k with hat on top left parenthesis 10 minus 3 right parenthesis
space space space space equals negative 17 straight i with hat on top space plus space 13 straight j with hat on top space plus space 7 straight k with hat on top
Hence comma 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 space equals space square root of 17 squared plus 13 squared plus 7 squared end root
space space space space space space space space equals open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar space equals space square root of 507
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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


open parentheses straight i with hat on top plus space 3 space straight j with hat on top plus space 9 space straight k with hat on top close parentheses space space straight x 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 0

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 1 3 9 row 3 cell negative straight lambda end cell straight mu end table close vertical bar space equals 0

straight i with hat on top left parenthesis 3 straight mu space plus space 9 straight lambda right parenthesis minus straight j with hat on top left parenthesis straight mu minus 27 right parenthesis space plus space straight k with hat on top left parenthesis negative straight lambda minus 9 right parenthesis space equals 0

3 straight mu space plus space 9 straight lambda space equals 0 space space..... space left parenthesis straight i right parenthesis

27 minus straight mu space equals 0 space space space....... left parenthesis ii right parenthesis

minus straight lambda minus 9 space equals 0 space...... space left parenthesis iii right parenthesis

By space equation space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis space

straight mu space equals space 27 space and space straight lambda space equals space minus 9

straight lambda comma space straight mu space value space satisfy space the space equation space left parenthesis straight i right parenthesis

so space space straight mu space equals space 27 comma space straight lambda space equals negative 9
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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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