Find the vector equation of the plane passing thrugh the points (2,5,-3),(-2,-3,5),(5,3,-3).

Explanation

The position vectors of the given points A, B, and C are `veca=(2hati+5hatj-3hatk), vecb=(-2hati-3hatj+5hatk)` and `vecc=(5hati+3hatj-3hatk)` respectively.
Then,the vector equation of the required plane is
`(vecr-veca).[(vecb-veca) xx (vecc - veca)]=0`.
Now, `(vecb-veca) = (-2-2)hati+(-3-5)hatj+(5+3)hatk=(-4hati-8hatj+8hatk)`. and `(vecc-veca)=(5-2)hati+(3-5)hatj+(-3+3)hatk=(3hati-2hatj+0hatk)`.
`therefore (vecb-veca) xx (vecc-veca) = |{:(hati,hatj,hatk),(-4,-8,8),(3,-2,0):}|`
`=(0+16)hatik-(0-24)hatj+(8+24)hatk`
`=(16hati+24hatj+32hatk)`.
`therefore` the required vector equation is
`(vecr-veca).(16hati+24hatj+32hatk)=0`
`rArr vecr.(16hati+24hatj+32hatk)=veca.(16hati+24hatj+32hatk)`
`=(2hati+5hatj-3hatk)/(16hati+24hatj+32hatk)`
`=(32+120-96)=56`.
Hence, the required vector equation of the plane is
`vecr.(16hati+24hatj+32hatk)=56 rArr vecr.(2hati+3hatj+4hatk)=7`.