Answer:
( xo , yo , zo ) = ( 1 , 0 , 1 )
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
- 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 )