NIMCET 2017 Question Paper with solutions

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Question : 100 of 120

Marks: +1, -0

If

ai↖{\^}+j↖{\^}+k

,

i↖{\^}+bj↖{\^}+k

,

i+j↖{\^}+ck↖{\^}

(a ≠ b ≠ c ≠ 1) are coplanar,< then the value of

1/{1-a}+1/{1-b}+1/{1-c}

is

- 1
$-1/2$
$1/2$
1

Solution:

If

ai↖{\^}+j↖{\^}+k↖{\^}

,

i↖{\^}+bj↖{\^}+k↖{\^}

,

i↖{\^}+j↖{\^}+ck↖{\^}

(a ≠ b ≠ c ≠ 1) are coplanar, then

|\table a,1,1;1,b,1;1,1,c|

= 0
=

|\table a,1-a,0;1,b-1,1-b;1,0,c-1|

= 0

(C_2→C_2-C_1)

(C_3→C_3-C_2)

= (1 - a) (1 - b) (1 - c)

|\table a/{1-a},1,0;1/{1-b},-1,1;1/{1-c},0,-1|

= 0
=

a/{1-a}

(1 - 0) -

1/{1-b}

( - 1 - 0) +

1/{1-c}

(1 - 0) = 0
⇒

a/{1-a}

+

1/{1-b}+1/{1-c}

= 0
⇒

a/{1-a}

+ 1 +

1/{1-b}+1/{1-c}

= 1
⇒

{a+1-a}/{1-a}

+

1/{1-b}+1/{1-c}

= 1
⇒

1/{1-a}+1/{1-b}+1/{1-c}

= 1

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