NIMCET 2015 Question Paper with solutions

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Question : 24 of 120

Marks: +1, -0

If

a↖{→}

=

i↖{\^}-k↖{\^}

,

b↖{→}

=

xi↖{\^}+j↖{\^}+(1-x)k↖{\^}

and

c↖{→}

=

yi↖{\^}+xj↖{\^}+(1+x-y)k↖{\^}

, then

[a↖{→}b↖{→}c↖{→}]

depends on;

Neither x nor y
Only x
Only y
Both x and y

Solution:

a=i↖{⋀}-k↖{⋀}

,

b=xi↖{⋀}+j↖{⋀}+(1-x)k↖{⋀}

and

c= yi↖{⋀}+xj↖{⋀}+(1+x-y)k↖{⋀}

Now

[abc]=|\table1,0,-1;x,1,1-x;y,x,1+x-y|

c_1→c_1+c_2=|\table0,0,-1;1,1,1-x;1+x,x,1+x-y|

Expand along

R_1

,

-1[x-1+x]=1

∴ Neither depends on x nor on y.

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