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Find the Vector and Cartesian Equations of the Plane Passing Through the Points (2, 2 –1), (3, 4, 2) and (7, 0, 6). Also Find the Vector

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प्रश्न

Find the vector and Cartesian equations of the plane passing through the points (2, 2 –1), (3, 4, 2) and (7, 0, 6). Also find the vector equation of a plane passing through (4, 3, 1) and parallel to the plane obtained above.

योग
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उत्तर

step 1
The given points are A(2, 2, –1), B(3, 4, 2) and C(7, 0, 6)
Let `vec"a" = 2hat"i" + 2hat"j" - hat"k"`
`vec"b" = 3hat"i" + 4hat"j" + 2hat"k"`
`vec"c" = 7hat"i" + 6hat"k"`
Hence the vector equation of the plane passing through the points
`(vec"r"- vec"a").(vec"AB" xx  vec"AC") = 0`

= `(vec"r" - vec"a").((vec"b" - vec"a") xx (vec"c" - vec"a")) = 0`

Now
`vec"b"- vec"a" = (3hat"i"+4hat"j" + 2hat"k")-(2hat"i"+ 2hat"j"-hat"k")`

⇒ `hat"i" + 2hat"j" + 3hat"k"`

`vec"c" - vec"a" = (7hat"i" + 6hat"k") - (2hat"i" + 2hat"j" - hat"k")`

= `5hat"i" - 2hat"j" + 7hat"k"`

So the required vector equation of plane is 
`[vec"r" - (2hat"i" + 2hat"j"  - hat"k")].[(hat"i" + 2hat"j" + 3hat"k") xx (5hat"i" - 2hat"j" + 7hat"k")] = 0`

Step 2
`(vec"b" - vec"a") xx (vec"c" - vec"a") = |(hat"i",hat"j",hat"k"),(1,2,3),(5,-2,7)|`

= `hat"i" (14 +6) -hat"j" (7 -15) + hat"k" (-2-10)`
= `20hat"i" + 8hat"j" - 12hat"k"`

⇒ `(vec"r" - (2hat"i" + 2hat"j" - hat"k")) . (20hat"i" + 8hat"j"-12hat"k") = 0`

`(vec"r" - (2hat"i" + 2hat"j" - hat"k")) . (5hat"i" + 2hat"j"- 3hat"k")  = 0`

`vecr . (5hat"i" + 2hat"j"- 3hat"k")  = (2hat"i" + 2hat"j" - hat"k"). (5hat"i" + 2hat"j", 3hat"k")`

`vecr . (5hat"i" + 2hat"j"- 3hat"k")  = 10 + 4 + 3`
`vecr . (5hat"i" + 2hat"j"-3hat"k")  = 17`
This is the required vector equation of the plane

Step 3
The Cartesian Equation of the plane passing through the three points is given as below-
5x + 2y − 3z − 17 = 0
This is the required cartesian equation of the plane.

The equation of plane parallel to 5x + 2y − 3z − 17 = 0 will be 5x + 2y − 3z + λ = 0
it passes through (4, 3, 1).
So, 5 × 4 + 2 × 3 − 3 × 1 + λ = 0
20 + 6 − 3 + λ = 0
So, λ = −23
so the equation of the plane will be
5x + 2y − 3z − 23 = 0
5x + 2y − 3z = 23
so the vector form of the equation of plane will be 
`vecr. (5hat"i" + 2hat"j" - 3hat"k") = 23`

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2018-2019 (March) 65/1/3
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