Question

Chap 6
vector

Answer 1

Find the vector equation and cartesian eqn of the line that is parallel to the vector -i+3j+2k and passes though the point P, position vector 2i-j-k.

Answer 2

Can you sketch the situation

Answer 3

@ganeshie8

Answer 4

the equation for a line in 3 dimensions is
r(t) = ro + t*v , where ro is a point on the line and v is a vector parallel to your line

Answer 5

point P on the line is (2,-1,-1)
and v is a scalar multiple of vector <-1, 3, 2>
r(t) = (2,-1,-1) + t* <-1,3,2>

Answer 6

you can use angular brackets for point (2,-1,-1) to turn it into a standard position vector , that is, a vector that starts at (0,0,0) and ends at (2,-1,-1)

Answer 7

so thats the vector equation.
the cartestian equation, we have
r(t) = < x(t), y(t), z(t) >
So
< x(t), y(t), z(t) > = <2,-1,-1> + t* <-1,3,2>
= <2 - t , -1 + 3t , -1 + 2t >
so cartesian equation is
x = 2- t
y = -1 + 3t
z = -1 + 2t
I think you can even eliminate t

(x-2)/-1 = (y + 1) /3 = (z+1)/2

Answer 8

but this is a bit odd, having a three way equality, but there is no way to eliminate the third variable

Answer 9

@perl

Answer 10

I got the solution.. but anyway thanks...
@perl

Answer 11

Can you sketch the situation for me
vector equation and cartesian eqn of the line that is parallel to the vector -i+3j+2k and passes though the point P, position vector 2i-j-k
on the x and y-axis
@perl

Answer 12

@hartnn

Answer 13

its in 3D so sketching might be a bit hard

Answer 14

http://prntscr.com/4v1qo2

Answer 15

Kk, thx
@ganeshie8