1
Question
Find the equation of the plane passing through the following points:
(1,1,1),(1,−1,2) and (−2,−2,2)
(1,1,1),(1,−1,2) and (−2,−2,2)
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Solution
Vector equation of a plane passing through three points with position vectors
→a,→b,→c is (→r−→a).[(→b−→a)×(→c−→a)]=0
Now, the plane passes through the points
A(1,1,1),B(1,−1,2) and C(−2,−2,2)→b−→a=^i−^j+2^k−^i−^j−^k=−2^j+^k
→c−→a=−2^i−2^j+2^k−^i−^j−^k=−3^i−3^j+^k(→b−→a)×(→c−→a)=∣∣
∣
∣∣^i^j^k0−21−3−31∣∣
∣
∣∣
=(−2+3)^i−(0+3)^j+(0−6)^k=^i−3^j−6^k.
(x^i+y^j+z^k−^i−^j−^k)(^i−3^j−6^k)=0⇒[(x−1)^i+(y−1)^j+(z−1)^k](^i−3^j−6^k)=0
⇒x−1−3y+3−6z+6=0⇒x−3y−6z+8=0 is the required equation of the plane.
Vector equation of a plane passing through three points with position vectors
→a,→b,→c is
(→r−→a).[(→b−→a)×(→c−→a)]=0
Now, the plane passes through the points
A(1,1,1),B(1,−1,2) and C(−2,−2,2)
→b−→a=^i−^j+2^k−^i−^j−^k=−2^j+^k
→c−→a=−2^i−2^j+2^k−^i−^j−^k=−3^i−3^j+^k
(→b−→a)×(→c−→a)=∣∣
∣
∣∣^i^j^k0−21−3−31∣∣
∣
∣∣
=(−2+3)^i−(0+3)^j+(0−6)^k
=^i−3^j−6^k.
(x^i+y^j+z^k−^i−^j−^k)(^i−3^j−6^k)=0
⇒[(x−1)^i+(y−1)^j+(z−1)^k](^i−3^j−6^k)=0
⇒x−1−3y+3−6z+6=0
⇒x−3y−6z+8=0 is the required equation of the plane.
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