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.(¯i−3^j+^k)=−1
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B
¯r.(−¯i+3^j−^k)=1
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C
¯r.(¯i−3^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^k10−1122∣∣
∣
∣∣=^i(0+2)−^j(2+1)+^k(2−0)=2^i−3^j+2^k⇒^nordirectionperpendiculartotheplaneneededis2^i−3^j+2^k√17
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^i−3^j+2^k)√17=0¯r.(¯¯¯¯¯2i−3^j+2^k)=(^i+^j).(2^i−3^j+2^k)¯r.(2^i−3^j+2^k)=2−3=−1∴¯r.(¯i−3^j+^k)=−1
This can also be written as ¯r.(−¯i+3^j−^k)=1
Therefore both a and b are correct.