Find the ratio in which the plane 2x-3y+z=8 divides the line segment joining the points A(3, -2,1) and B(1, 4, -3). Also find the point of intersection of the line and the plane.

Answer

N/a

Explanation

Let plane divides the line segment AB at point R (x,y,z) and R divides AB in the ratio `lamda : 1`.
`therefore x=(lamda+3)/(lamda+1)`
`y=(4lamda-2)/(lamda+1)`
`z=(-3lamda+1)/(lamda+1)`
Now, the point R(x,y,z), lies on the plane 2x-3y+z=8
`therefore (2(lamda+3))/(lamda+1)-(3(4lamda-2))/(lamda+1)+(-3 lamda+1)/(lambda+1)=8`
` implies 2 lamda+6-12 lamda +6-3lamda+1=8lamda+8`
`implies 21 lambda=5`
`implies lamda=(5)/(21)=5 : 21`
Therefore, the required ratio=5 : 21. and the co-ordinates of R
`(((5)/(21)+3)/((5)/(21)+1),((20)/(21)-2)/((5)/(21)+1),((-15)/(21)+1)/((5)/(21)+1))`
`((68)/(26).(-22)/(26),(6)/(26))`
`=((34)/(13),(-11)/(13),(3)/(13))`