1
Question
Four points given by position vectors 2→a+3→b−→c, →a−2→b+3→c, 3→a+4→b−2→c and →a−6→b+6→c are coplanar, where →a,→b and →c are non-coplanar vectors.
Four points given by position vectors 2→a+3→b−→c, →a−2→b+3→c, 3→a+4→b−2→c and →a−6→b+6→c are coplanar, where →a,→b and →c are non-coplanar vectors.
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Solution
The correct option is A True
Let four points be A, B, C and D respectively and O be the origin, then
O→A=2→a+3→b−→cO→B=→a−2→b+3→cO→C=3→a+4→b−2→cO→D=→a−6→b+6→c⎤⎥
⎥
⎥
⎥
⎥⎦ Just because these 4 points are position vectors with respect to origin.
Now let us form 3 vectors out of these 4 points. Why? Because for 3 vectors we know the working rule. We just have to check their linear dependence and if they are linearly dependent, then they are coplanar. Else not.
Let α=A→B=O→B−O→A (Triangle law for vector addition)
=−→a−5→b+4→c
β=A→C=O→C−O→A
=→a+→b+→c
γ=A→D=O→D−O→A=−→a−9→b+7→c
For linear dependence
Let →α=x→β+y→γ
−→a−5→b+4→c=x(→a+→b+→c)+y(−→a−9→b+7→c)
=(x−y)→a+(x−9y)→b−(x−7y)→c
Equating the coefficients of →a,→b and →c. We get
x – y = -1 …(1)
x – 9y = -5 …(2)
-x + 7y = 4 …(3)
Solving (1) and (2) we get y=12
x=−12
and these values satisfy equation 3 as well. So system of equations is consistent with a non-zero solution. So α,β and γ are coplanar. Hence points A, B, C and D are coplanar. (Because −−→AB,−−→AC and −−→AD lie in one plane, so −−→OA, −−→OB −−→OC and −−→OD all must lie in that plane).
Let four points be A, B, C and D respectively and O be the origin, then
O→A=2→a+3→b−→cO→B=→a−2→b+3→cO→C=3→a+4→b−2→cO→D=→a−6→b+6→c⎤⎥ ⎥ ⎥ ⎥ ⎥⎦ Just because these 4 points are position vectors with respect to origin.
Now let us form 3 vectors out of these 4 points. Why? Because for 3 vectors we know the working rule. We just have to check their linear dependence and if they are linearly dependent, then they are coplanar. Else not.
Let α=A→B=O→B−O→A (Triangle law for vector addition)
=−→a−5→b+4→c
β=A→C=O→C−O→A
=→a+→b+→c
γ=A→D=O→D−O→A=−→a−9→b+7→c
For linear dependence
Let →α=x→β+y→γ
−→a−5→b+4→c=x(→a+→b+→c)+y(−→a−9→b+7→c)
=(x−y)→a+(x−9y)→b−(x−7y)→c
Equating the coefficients of →a,→b and →c. We get
x – y = -1 …(1)
x – 9y = -5 …(2)
-x + 7y = 4 …(3)
Solving (1) and (2) we get y=12
x=−12
and these values satisfy equation 3 as well. So system of equations is consistent with a non-zero solution. So α,β and γ are coplanar. Hence points A, B, C and D are coplanar. (Because −−→AB,−−→AC and −−→AD lie in one plane, so −−→OA, −−→OB −−→OC and −−→OD all must lie in that plane).
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