The cartesian equation of the line through the points (1,2,3) which is parallel to the vector 4 ̂i + 2 ̂j + 6 ̂k is given by-
Answer & explanation
Correct answer: option 4
We have vector a = ̂i + 2 ̂j + 3 ̂k
and vector b = 4 ̂i + 2 ̂j + 6 ̂k
Therefore vector equation of the line is given by-
vector r = a+λb
⇒ r = ( ̂i + 2 ̂j + 3 ̂k)+λ (4 ̂i + 2 ̂j + 6 ̂k)
⇒ r = (1+ 4λ) ̂i + (2+2λ) ̂j + (3+ 6λ) ̂k
⇒ x ̂i + y ̂j +z ̂k = (1+ 4λ) ̂i + (2+2λ) ̂j + (3+ 6λ) ̂k
Eliminating λ, we get
(x-1)/4 = (y-2)/2 = (z-3)/6