Question

Find the coordinates of the point where the line through (3, ­−4, −5) and (2, − 3, 1) crosses the plane 2 x + y + z = 7).

Answer 1

The given plane points are ( −3,−4,−5 ) and ( 2,−3,1 ).

We know that the equation of the line passing through the points, ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ), is given by,

x− x 1 x 2 − x 1 = y− y 1 y 2 − y 1 = z− z 1 z 2 − z 1 =k (1)

Substitute the values of x 1 , y 1 , z 1 and x 2 , y 2 , z 2 in equation (1).

x−3 2−3 = y−( −4 ) −3−( −4 ) = z−( −5 ) 1−( −5 ) =k x−3 −1 = y+4 1 = z+5 6 =k

x−3=−k x=−k+3

y+4=k y=k−4

z+5=6k z=6k−5

Thus, a point on the given line is of the form, ( 3−k,k−4,6k−5 ).

The given point lies on the plane whose equation is given by,

2x+y+z=7(2)

Substitute values of x,y and z in equation (2),

2( 3−k )+( k−4 )+( 6k−5 )=7 5k−3=7 k=2

Put the value of k in the given point,

x=3−k =3−2 =1

y=k−4 =2−4 =−2

z=6k−5 =12−5 =7

Thus, the required point is ( 1,−2,7 ).