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Question

Given 3 points given by position vectors ¯a=^i+^j,¯b=^j+^k,¯c=^i^j^k. Find the plane passing through these 3 points.

A
¯r.(¯i3^j+^k)=1
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B
¯r.(¯i+3^j^k)=1
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C
¯r.(¯i3^j+^k)=2
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D
¯r.(¯i+3^j^k)=2
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Solution

The correct option is B ¯r.(¯i+3^j^k)=1
First we have to find the vector perpendicular to the plane. It will be along (¯a¯b)×(¯b¯c)
(¯a¯b)×(¯b¯c)=((^i+^j)(^j+^k)×(^j+^k)(^i^j^k))=(^i^k)×(^i+2^j+2^k)=∣ ∣ ∣^i^j^k101122∣ ∣ ∣=^i(0+2)^j(2+1)+^k(20)=2^i3^j+2^k^n or direction perpendicular to the plane needed is 2^i3^j+2^k17
Without loss of generality we will choose ¯a to be the point on the plane. Therefore equation of the plane will be (¯r¯a).^n=0
(¯r(^i+^j)).^n=0(¯r(^i+^j)).(2^i3^j+2^k)17=0¯r.(¯¯¯¯¯2i3^j+2^k)=(^i+^j).(2^i3^j+2^k)¯r.(2^i3^j+2^k)=23=1¯r.(¯i3^j+^k)=1
This can also be written as ¯r.(¯i+3^j^k)=1
Therefore both a and b are correct.


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