If the vector a i+j+k, i+b j+k,i+j+c k(a, b, c \\neq1) are coplanar then \frac{1}{1-a}+\frac{1}{1-b}+{1}/_{}{1-c}=

NIMCET 2021 Question Paper with solutions

Question : 29 of 120

Marks: +1, -0

If the vector

a i+j+k, i+b j+k,i+j+c k(a, b, c \≠1)

are coplanar then

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

Solution:

For coplanar vector

[{{a}↖{→}}, {b} ↖{→},{{c}↖{→}}]=0

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

{R}_{2} \→{R}_{2}-{R}_{1}

and

{R}_{3} \→ {R}_{3}-{R}_{1}

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

Expand

R_{1}

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

+1(a-1)(b-1)=0

a(b-1)(c-1)+(a-1)(c-1)

+(a-1)(b-1)=0

Divide by

({a}-1)({b}-1)({c}-1)

{1}/{ a-1}+{1}/{ b-1}+{1}/_{}{ c-1}=0

Take - ve sign common

{1}/{1-a}+{1}/{1-b}+{1}/_{}{1-c}=0

Add both side 1

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

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

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