Find the vector equation of the line passing through (1, 2, 3) and parallel to each of the planes r ( i j + 2 k) = 5 and r (3 i + j + k) = 6. Also find the point of intersection of the line thus obtained with the plane r (2 i + j + k) = 4.
Answers
Vector equation of line is :
(i + 2j + 3k)+k( -i + 5j -2k )
Coordinates of point of intersection of line are ( 4, -13 ,9)
•let direction ratios of required line
be a, b, c
•Since , Required line is parallel to
given plane
plane 1 : r ( i + j + 2 k) = 5
plane 2 : r (3 i + j + k) = 6
•dot product of direction ratios wil
be zero
a + b + 2c = 0
3a + b + c = 0
•using cross multiplication method
a/[(1)-(2)] = -b/[(1)-(6)] = c/[(1)-(3)]
a/(-1) = -b/(-5) = c/(-2)
a/(-1) = b/(5) = c/(-2)
•Required direction ratios are
a = -1 b = 5 c = -2
•Cartesian equation of line is :
(x-1)/(-1) = (y-2)/(5) = (z-3)/(-2)
•Vector equation of line is :
(i + 2j + 3k)+k( -i + 5j -2k )
•for point of intersection with plane
p : r (2 i + j + k) = 4
i.e. 2x+y+z = 4
make the general points of the intersecting line
•(x-1)/(-1) = (y-2)/(5) = (z-3)/(-2) = k
(x-1)/(-1) = k
x = 1-k
(y-2)/(5) =k
y = 2 + 5k
(z-3)/(-2) = k
z = 3-2k
•Now, these point should pass through plane for intersection
•putting x , y , z in plane
2(1-k) + 5k+2 + 3-2k = 4
2 - 2k + 5k+2 + 3-2k = 4
k = -3
•Now
x = 1-k = 1-(-3) = 4
y = 5k +2 = 5(-3) +2 = -13
z = 3-2k = 3 - 2(-3) = 9
•Coordinates of point of intersection are ( 4, -13 ,9)