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Question

The equation of a plane passing through ¯r.(^i+^j+^k)=1 and ¯r.(2^i+^k)=2 can be given by

A
¯r.((1+2λ)^i+^j+(1+λ)^k)=1+2λ
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B
¯r.((λ+2)^i+λ^j+(λ+1)^k)=λ+2
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C
¯r.((3λ^i+2λ^j+2λ^k)=3)=3λ
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D
¯r.(^i+^j+^k)=1
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Solution

The correct option is D ¯r.(^i+^j+^k)=1
We already know equation of any plane passing through intersection of ¯r.^n1=d1 and ¯r.^n2=d2 will be ¯r(^n1+λ^n2)=d1+λd2
Therefore, ¯r.(^i+^j+^k+λ(2^i+^k))=1+2λ
¯r.((1+2λ)^i+^j+(1+λ)^k)=1+2λ
Option a:- Satisfies.

But we can do it the other way too i.e,

¯r.(λ^n1+^n2)=λd1+d2 which gives,
¯r.((λ+2)^i)+λ^j+(λ+1)^k)=λ+2

Option C is nothing but ¯r.λ(3^i+2^j+2^k)=3λ
For λ0 this can be written as,
λ(¯r.((^i+^j+^k+1(2^i+^k)))=(1+2)λ
Which is a linear combination of 2 planes.
Option d can be arrived easily by substituting λ=0 in r.(n1+λ^n2)=d2+λd2
Answer is a,b,c,d




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