Question

Show that the points (-1,4,-3) , (3,2,-5) , (-3,8,-5) and (-3,2,1) are coplanar?

Answer 1

They are not coplanar . Still i could always be wrong !

Explanation:

using the four points , we can create 3 vectors with one of the point as common origin.

A(-1,4,-3) ; B(3,2,-5); C(-3,8,-5); D(-3,2,1)

i take A as common origin (doesn't matter which one you take) .
So i create 3 vectors AB , AC , and AD.
`vec (AB)` = <3-(-1),2-4,-5-(-3)> = <4,-2,-2>
`vec(AC)` = <-3-(-1),8-4,-5-(-3)> = <-2 , 4, -2>
`vec(AD)` = <-3-(-1),2-4,1-(-3)> = <4,-2,4>

Take take the cross product of any of the two vectors(this will give you a resulting vector which is normal to the plane ) and perform the dot product of the result with the remaining vector .

If you get 0 as the result , then the four points are coplanar

I'm taking cross product of AB and AC
`vec(AB) X vec(AC)` = <-12,-4,12> . I have skipped the calculations.

now dot product this result with AD
`vec(res) . vec(AD)` = <-12,-4,12> . <4,-2,4>
= 8 `cancel= 0`.
Therefore they are not coplanar .
I can always be wrong !

Answer 2

`" "A-=(-1,+4,-3)`
`" "B-=(+3,+2,-5)`
`" "C-=(-3,+8,-5)`
`" "D-=(-3,+2,+1)`
lie in a single plane

Explanation:

Let
`" "A-=(-1,+4,-3)`
`" "B-=(+3,+2,-5)`
`" "vec(AB)=(3-(-1))hati+(2-4)hatj+(-5-(-3))hatk`
`vec(AB)=4hati-2hatj-2hatk`

`" "A-=(-1,+4,-3)`
`" "C-=(-3,+8,-5)`
`" "vec(AC)=(-3-(-1))hati+(8-4)hatj+(-5-(-3))hatk`
`vec(AC)=-2hati-2hatj+4hatk`

`" "A-=(-1,+4,-3)`
`" "D-=(-3,+2,+1)`
`" "vec(AD)=(-3-(-1))hati+(2-4)hatj+(1-(-3))hatk`
`vec(AD)=-2hati-2hatj+4hatk`

`row1=+4,-2,-2`
`row2=-2,-2,+4`
`row3=-2,-2,+4`
Since two rows in the determinant are same
determinant vanishes to zero.
Thus,
`" "A-=(-1,+4,-3)`
`" "B-=(+3,+2,-5)`
`" "C-=(-3,+8,-5)`
`" "D-=(-3,+2,+1)`
lie in a single plane