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- 1. Let L be the line in R^3 passing through the point (2, 2, 5) and parallel to the vector (2,-1, 1).Let S be the the line segment consisting of the points on L whose coordinates are all nonnegative (that is, S is the intersection of L with the first octant). Find the vector equation of S. Solution The vector equation of the line will be r = 2i +2j + 5k + t*(2i -j +k ) r = (2+2t) i + (2-t) j + (5+t) k Now we require the part which lies in the first octant i.e all co-ordinates are >=0 2+2t >=0 & 2-t >= 0 & 5+t >=0 This gives us that t>=-1 &t<=2 and t>=-5 respectively. Taking their intersection you'll get that -1<=t<=5 So the equation will be the same but we only impose a condition on t that -1<=t<=5