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Question
The coordinates of coplanar points A,B,C,D are (2−x,2,2),(2,2−y,2),(2,2,2−z) and (1,1,1) respectively, then
The coordinates of coplanar points A,B,C,D are (2−x,2,2),(2,2−y,2),(2,2,2−z) and (1,1,1) respectively, then
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Solution
The correct option is C 1x+1y+1z=1
−−→AB=−−→OB−−−→OA
={2^i+(2−y)^j+2^k}−{(2−x)^i+2^j+2^k}=x^i−y^j
−−→AC=−−→OC−−−→OA
={2^i+2^j+(2−z)^k}−{(2−x)^i+2^j+2^k}=x^i−z^k
−−→AD=−−→OD−−−→OA={^i+^j+^k}−{(2−x)^i+2^j+2^k}
=(x−1)^i−^j−^k
As these vectors are coplanar, ∣∣
∣∣x−y0x0−zx−1−1−1∣∣
∣∣=0
On simplification, we get 1x+1y+1z=1
Hence, option 'A' is correct.
−−→AB=−−→OB−−−→OA
={2^i+(2−y)^j+2^k}−{(2−x)^i+2^j+2^k}=x^i−y^j
−−→AC=−−→OC−−−→OA
={2^i+2^j+(2−z)^k}−{(2−x)^i+2^j+2^k}=x^i−z^k
−−→AD=−−→OD−−−→OA={^i+^j+^k}−{(2−x)^i+2^j+2^k}
=(x−1)^i−^j−^k
As these vectors are coplanar, ∣∣ ∣∣x−y0x0−zx−1−1−1∣∣ ∣∣=0
On simplification, we get 1x+1y+1z=1
Hence, option 'A' is correct.
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