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Question
The vector equation of the plane through the point →i+2→j−→k and ⊥ to the line of intersection of the plane →r.(3→i−→j+→k)=1 and →r.(→i+4→j−2→k)=2 is
The vector equation of the plane through the point →i+2→j−→k and ⊥ to the line of intersection of the plane →r.(3→i−→j+→k)=1 and →r.(→i+4→j−2→k)=2 is
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Solution
The correct option is C →r.(2→i−7→j−13k)=1
The line of inetrsection of the planes →r.(3→i−→j+→k)=1 and →r.(→i+4→j−2→k)=2 is common to both the planes.Therefore, it is ⊥ to normals to the two planes i.e. →n1=3→i−→j+→k and →n2=→i+4→j−2→k.Hence, it is parallel to the vector →n1×→n2=−2→i+7→j+13→k.Thus, we have to find the equation of the plane passing through →a=→i+2→j−→k and normal to the vector →n=→n1×→n2.The equation of the required plane is(→r−→a).→n⇒→r.→n=→a.→n⇒→r.(−2→i+7→j+13→k)=(→i+2→j+→k).(−2→i+7→j+13→k)⇒→r.(2→i−7→j−13→k)=1
The line of inetrsection of the planes →r.(3→i−→j+→k)=1 and →r.(→i+4→j−2→k)=2 is common to both the planes.
Therefore, it is ⊥ to normals to the two planes i.e. →n1=3→i−→j+→k and →n2=→i+4→j−2→k.
Hence, it is parallel to the vector →n1×→n2=−2→i+7→j+13→k.
Thus, we have to find the equation of the plane passing through →a=→i+2→j−→k and normal to the vector →n=→n1×→n2.
The equation of the required plane is
(→r−→a).→n⇒→r.→n=→a.→n⇒→r.(−2→i+7→j+13→k)=(→i+2→j+→k).(−2→i+7→j+13→k)⇒→r.(2→i−7→j−13→k)=1
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