Find the point in which the line x=1-t, y=3t, z=1+t; meet the plane 2x-y+3z=6

Answer 1

x = 1 + 2t

y = 2 - 3t

z = -5 + t

Also, we know that:

2x + 5y - 3z = 6

Substituting for x, y and z in the that equation gives:

2 * (1 + 2t ) + 5 * ( 2 - 3t ) - 3 * ( -5 + t ) = 6

2 + 4t + 10 - 15t + 15 - 3t = 6

4t - 15t - 3t = 6 - 2 - 10 - 15

- 14t = - 21 Divide both sides with -14

t = - 21 / -14

t = ( 7 * 3 ) / ( 7 * 2 )

t = 3 / 2

x = 1 - t

x = 2 / 2 - 3 / 2

x = - 1 / 2

y = 3 t

y = 3 * 3 / 2

y = 9 / 2

Answer 2

x = 1 + 2t

y = 2 - 3t

z = -5 + t

Also, we know that:

2x + 5y - 3z = 6

Substituting for x, y and z in the that equation gives:

2 * (1 + 2t ) + 5 * ( 2 - 3t ) - 3 * ( -5 + t ) = 6

2 + 4t + 10 - 15t + 15 - 3t = 6

4t - 15t - 3t = 6 - 2 - 10 - 15

- 14t = - 21 Divide both sides with -14

t = - 21 / -14

t = ( 7 * 3 ) / ( 7 * 2 )

t = 3 / 2

x = 1 - t

x = 2 / 2 - 3 / 2

x = - 1 / 2

y = 3 t

y = 3 * 3 / 2

y = 9 / 2

z = 1 + t

z = 2 / 2 + 3 / 2

z = 5 / 2

Coordinate of point:

( - 1 / 2 , 9 / 2 , 5 / 2 )