How do you find the parametric equation of a line in R3 that passes through P(1,1,1) and is orthogonal to both the lines x=3-t, y= -1+4t, z = -2t and x=2+2t, y=1+t, z=-2+3t where t is any real number?

1 Answer
Dec 28, 2016

The answer is #x=1+14s ; y=1-s ;z=1-9s# ; #s in RR#

Explanation:

To find a line orthogonal to 2 other lines, we must perfom a cross product

The first line #L1# is

#x=3-t #

#y=-1+4t#

#z=-2t#

The vector parallel to line #L1# is #vecL_1=〈-1,4,-2〉#

The second line #L2# is

#x=2+2t#

#y=1+t#

#z=-2+3t#

The vector parallel to line #L2# is #vecL_2=〈2,1,3〉#

The vector orthogonal to #vecL_1# and #vecL_2=# is given by the cross product

# | (hati,hatj,hatk), (-1,4,-2), (2,1,3) | #

#=hati| (4,-2), (1,3) | -hatj| (-1,-2), (2,3) | +hatk| (-1,4), (2,1) | #

#=hati(12+2)-hatj(-3+4)+hatk(-1-8)#

#=〈14,-1,-9〉#

Verification by doing a dot product

#〈14,-1,-9〉.〈-1,4,-2〉=-14-4+18=0#

#〈14,-1,-9〉.〈2,1,3〉=28-1-27=0#

The equation of the line orthogonal to #L_1# and #L_2# and passing through #P(1,1,1)# is

#x=1+14s#

#y=1-s#

#z=1-9s#