Given 3 points given by position vectors ¯a,¯b and ¯c. The plane which passes through these 3 points can be given by
A
(¯r−¯a).((¯a+¯b)×(¯b+¯c))=0
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B
(¯r−¯a).((¯a+¯b)×(¯b−¯c))=0
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C
(¯r−¯b).((¯a−¯b)×(¯b−¯c))=0
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D
(¯r−¯a).((¯a−¯b)×(¯b−¯c))=0
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Solution
The correct option is D(¯r−¯a).((¯a−¯b)×(¯b−¯c))=0 For constructing a plane, one of the ways is to have a point through which the plane passes and perpendicular vector to the plane. Here we already have 3 points. We just have to calculate the direction vector. For that if we can get 2 vectors parallel to the plane we can get the perpendicular vector by cross product. If you see closely (¯a−¯b),(¯b−¯c) and (¯c−¯a) are parallel to the plane we want. So (¯a−¯b)×(¯b−¯c) serves as a direction vector. Now we have a point (either ¯aor¯bor¯c)and a direction vector perpendicular to the plane , therefore equation of the required plane can be given by (¯r−¯a).((¯a−¯b)×(¯b−¯c))=0or(¯r−¯b).((¯a−¯b)×(¯b−¯c))=0or(¯r−¯c).((¯a−¯b)×(¯b−¯c))=0. Therefore options c and d are correct.