Question

Find the line through (3, 1, −2) that intersects and is perpendicular to the line x = −1 + t, y = −2 + t, z = −1 + t. (HINT: If (x0, y0, z0) is the point of intersection, find its coordinates. Enter your answers as a comma-separated list of equations.)

Answer 1

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  

- 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 )