vector equation of a line?

Question

i really am so stuck with how to do this, im reading through my notes and just getting lost in all the different letters. somebody tell me how to do this kind of question please!!

heres an example of what i currently cant do:

find vector equation of line L which passes through (1,0,2) in the direction of (-1,4,2)

no way, so THATS all im being asked to do . . . cheers.

Answer 1

r , a and b are vectors.
µ is a scalar.
r = a + µ (b - a)
r = (1, 0, 2) + µ(-2 , 4 , 0) is vector equation of required line.

Answer 2

Vectors for AB and CD are ok. enable X be the ingredient of intersection of strains AB and CD. So we choose place vector of X relative to the beginning, i.e. vector OX considering that X is on line AB, then vector AX is scalar distinctive of vector AB AX = m (6v - 10u) OX = OA + AX = 10u + m (6v - 10u) considering that X is on line CD, then vector CX is scalar distinctive of vector CD CX = n (12v + 2u) OX = OC + CX = -2u + n (12v + 2u) OA + AX = OX = OC + CX 10u + m (6v - 10u) = -2u + n (12v + 2u) (10 - 10m) u + (6m) v = (-2 + 2n) u + (12n) v Matching coefficients, we get: 6m = 12n 10 - 10m = -2 + 2n From first equation: 2n = m From 2d equation: 2n = 12 - 10m m = 12 - 10m 11m = 12 m = 12/11 2n = m 2n = 12/11 n = 6/11 OX = OA + AX = 10u + m (6v - 10u) OX = 10u + 12/11 (6v - 10u) OX = 10u + seventy two/11 v - a hundred and twenty/11 u OX = seventy two/11 v - 10/11 u OX = OC + CX = -2u + n (12v + 2u) OX = -2u + 6/11 (12v + 2u) OX = -2u + seventy two/11 v + 12/11 u OX = seventy two/11 v - 10/11 u

Answer 3

Would you believe that it's just the offset plus multiples of the direction vector?

L(x,y,z) = (1,0,2) + t*(-1,4,2)