Respuesta :
Answer:
( xo , yo , zo ) = ( 1 , 0 , 1 )
Step-by-step explanation:
Given:-
- A line passing through point (3, 1, −2) intersects and is perpendicular to line with coordinates:
         x = −1 + t, y = −2 + t, z = −1 + t  .... t = arbitrary parameter.
Find:-
The coordinates for point of intersection.
Solution:-
- The line that passes through point (3, 1, −2) = ( a, b , c ) and an a arbitrary point on the given line have the following direction vector d2 :
         d2 = ( x2 , y2 , z2 )
         x2 = a - ( x ) = 3 - ( -1 + t ) = 4 - t
         y2 = b - ( y ) = 1 - ( -2 + t ) = 3 - t
         z2 = c - ( z )  = -2 - ( -1 + t ) = -1 - t
        d2 = (  4 - t , 3 - t , -1 - t )
- The direction vector d1 of the given line is:
         d1 = ( x1 , y1 , z1 )
         x1 = 1
         y1 = 1
         z1 = 1
         d1 = ( 1 , 1 , 1 )
- The dot product of two orthogonal vectors is always equal to zero:
         d1.d2 = 0
         (  4 - t , 3 - t , -1 - t ) . ( 1 , 1 , 1 ) = 0
- Solve for parameter (t):
         (4 - t) + (3 - t) + (-1 - t) = 0
         6 -3t = 0
         t = 2 Â
- The coordinates of the point of intersections can be evaluated by substituting the value of "t" into the given equation of line:
         xo ( t = 2) = - 1 + 2 = 1
         yo ( t = 2) = - 2 + 2 = 0
         zo ( t = 2) = - 1 + 2 = 1
- The coordinates are:
         ( xo , yo , zo ) = ( 1 , 0 , 1 )
