Find the equation of the line passing through (1,2,-4) and parallel to the lines r=i+2j-4k + lamda(2i+3j+6k) , r=i-3j+5k+ u(i+j-k) and also find the perpendicular distance of this plane from the origin?
Asked by Mrinal | 13th Mar, 2013, 09:19: PM
Please note, that we would be required to find the equation of a plane and not the line.
Since, the plane needs to be parallel to both the lines r1 and r2, so it means that a vector perpendicular to plane would be perpendicular to both the lines.
Let n be the perpendicular vector
so, n = (2i+3j+6k) x (i+j-k) [ cross product of the direction vectors of the 2 lines]
n = -9i + 8j - k
hence the equation of the plane: n. (r - r0) = 0
i.e. (-9i + 8j - k).(r - (i+2j-4k)) = 0
r. (-9i + 8j - k) - (-9+16+4) = 0
r. (-9i + 8j - k) = 11
Hence,the perpendicular direction is found by dividing the equation of the plane by the length of the normal vector. Length of normal vector = sqrt(92 + 82+12) = sqrt 146
So, the perpendicular distance from origin = 11/sqr(146)
Answered by | 14th Mar, 2013, 07:13: AM
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