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A.)
Find the distance each point and the line given by each set of parametric equations.
Hence the distance of given point from the line is 2.96 units
Given:
The coordinates of points Q is (1,5,–2) . The parametric equation of line is
x=–2+4ty=3z=1+t
Explanation:
Consider the
a×b=|ijkx1y1z1x2y2z2| =i[y1(z2)–y2(z1)]–j[x1(z2)–x2(z1)]+k[x1(y2)–x2(y1)]
Distance between a point Q and line in space with direction vector u is
D=∥PQ×u∥∥u∥
ThecoordinatesofpointsQis (1,5,–2) .The parametric equation of line is
x=–2+4ty=3z=1+t
Directionvector of line is u=〈4,0,–1〉 . Now the coordinates of a general point on the line will be
x=4 λ –2y=3z=1– λ
Therefore coordinates of point P is (4 λ–2,3,1–λ).
Hence vector ¯PQ is
¯PQ= (4 λ–2–1,3–5,1–λ+2) =(4 λ–3,–2,3–λ)
Now cross product of two vectors and is
¯PQ×u=[ijk4λ−3−23−λ40−1]=i(2−0)−j(−4λ+3−12+4λ)+k(0+8)=i(2)−j(−9)+k(8)=〈2,9,8〉
Therefore,
D=∥¯PQ×u∥∥u∥=√22+92+82√42+(−1)2=2.96
Hence the distance of given point from the line is 2.96 units.
B.)
The distance of given point from the line PQ is 0.745 units
Given:
The coordinates of points Q is (1,–2,4) . The parametric equation of line is
x=2ty=–3+tz=2+2t
Explanation:
Consider the vectors in space a=x1i+y1j+z1k and b=x2i+y2j+z2k then the cross product of a=x1i+y1j+z1k and b=x2i+y2j+z2k will be the vector
a×b=|ijkx1y1z1x2y2z2| =i[y1(z2)–y2(z1)]–j[x1(z2)–x2(z1)]+k[x1(y2)–x2(y1)]
Distance between a point Q and line in space with direction vector u is
D=∥PQ×u∥∥u∥
ThecoordinatesofpointsQis (1,–2,4) .Theparametricequationoflineis
x=2ty=–3+tz=2+2t
Directionvectoroflineis u= 〈2,1,2〉 .Nowthecoordinatesofageneralpointonthelinewillbe
x=2 λy=–3+ λz=2+2 λ
ThereforecoordinatesofpointPis (2 λ,–3+λ,2+2λ).
Hencevector ¯PQ is
¯PQ= 2 λ–1,–3+λ+2,2+2λ–4 =2 λ–1,–1+λ,–2+2 λ
Nowcrossproductoftwovectorsandis
¯PQ×u=[ijk2λ−1−1+λ−2+2λ212]=i(–2+2λ+2–2λ)–j(4λ–2+4 –4λ)+k(2 λ–1+2–2λ) =i(0)–j(–2)+k(1) =〈0,–2,1〉
Therefore,
D=∥¯PQ×u∥∥u∥=√−22+12√22+(1)2+22=0.745
Hencethedistanceofgivenpointfromtheline PQis0.745units.
Chapter 11 Solutions
EBK PRECALCULUS W/LIMITS
