Equation of the plane through three points A,B,C with position vectors −6→i+3→j+2→k,3→i−2→j+4→k,5→i+7→j+3→k is
A
→r⋅(→i−→j−7→k)+23=0
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B
→r⋅(→i+→j+7→k)=23
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C
→r⋅(→i+→j−7→k)+23=0
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D
→r⋅(→i−→j−7→k)=23
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Solution
The correct option is A→r⋅(→i−→j−7→k)+23=0 We have →A=−6→i+3→j+2→k,→B=3→i−2→j+4→k,→C=5→i+7→j+3→k Thus →AB=9→i−5→j+2→k,→AC=11→i+4→j+→k Normal vector perpendicular to plane ABC is →n=→AB×→AC ⇒→n=∣∣
∣
∣∣→i→j→k9−521141∣∣
∣
∣∣=−13→i+13→j+91→k=13(−→i+→j+7→k) Thus equation of plane ABC is (→r−→A)⋅→n=0 ⇒(x+6)(−1)+(y−3)(1)+(z−2)(7)=0 ⇒x+6−y+3−7z+14=0 ⇒(x−y−7z)+23=0 ⇒→r⋅(→i−→j−7→k)+23=0 Hence, option 'A' is correct choice.