Question 1301c

Vectors II: Lines and Planes
Cartesian Equation of a Line

Question

Given the line l{l} with cartesian equation
x+1−3=z−55,y=−5,\frac{x + 1}{-3} = \frac{z - 5}{5}, y = - 5,
convert the equation of l{l} into vector form.

Attempt

l:{l: } r=a+λd,  λ∈R.{\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}, \; \lambda \in \mathbb{R}.}
a={\mathbf{a}=}
({\Biggl(}
),{\Biggr), \quad}
d={\mathbf{d}=}
({\Biggl(}
){\Biggr)}

Answer

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