Determine the vector equation of each line:
a) Direction vector [1,-2] and through P(-4,11)
b) Normal vector of [5,-2] and through P (-1,2)
c) Parallel to x + 6y = 1 and through P (2, -5)
d) Parallel to the z-axis and through P(-4,3,7)
e) Parallel to [x,y,z] = [1,3,-2] = t[5,2,1] and through P(2,0,1)

Question

Determine the vector equation of each line:
a) Direction vector [1,-2] and through P(-4,11)
b) Normal vector of [5,-2] and through P (-1,2)
c) Parallel to x + 6y = 1 and through P (2, -5)
d) Parallel to the z-axis and through P(-4,3,7)
e) Parallel to [x,y,z] = [1,3,-2] = t[5,2,1] and through P(2,0,1)

anonymous · Answer

$$r=i-2j+\lambda \left( -4i+11j \right)$$

anonymous · Answer

That is just for a) correct? But thank you!

anonymous · Answer

eq of any line parallel to x+6y=1 is
x+6y=c
it passes through (2,-5)
2+6*-5=c
c=2-30=-28
hence x+6y=-28

anonymous · Answer

$$\left\{ r-\left( -i+2j \right) \right\}.\left( 5i-2j \right)=0$$
$$r.\left( 5i-2j \right)+5+4=0$$
$$r.\left( 5i-2j \right)+9=0$$

anonymous · Answer

For x+6y=c is the answer simply  x+6y=-28 for a vector equation?

anonymous · Answer

rest in the morning .good night.

anonymous · Answer

thank you so much for your help!