Step 1

Given:

The given point is \(\displaystyle{\left(−{1},\ {0},\ {8}\right)}\) and the vector or line is \(\displaystyle{v}={3}{i}+{4}{j}-{8}{k}\)

To determine: The sets of parametric equations and symmetric equations of the line that passes through the given point and is parallel to the given vector or line.

Step 2

a) The parametric equations for a line passing through \(\displaystyle{\left({x}_{{{0}}},\ {y}_{{{0}}},\ {z}_{{{0}}}\right)}\) and parallel to the vector \(\displaystyle{v}={a}{i}+{b}{j}+{c}{k}\) are

\(\displaystyle{x}={x}_{{{0}}}+{a}{t},\ {y}={y}_{{{0}}}+{b}{t},\ {z}={z}_{{{0}}}+{c}{t}\)

The required parametric equations are

\(\displaystyle{x}=-{1}+{3}{t},\ {y}={4}{t},\ {z}={8}-{8}{t}\)

Step 3

b) The parametric equations are

\(\displaystyle{x}=-{1}+{3}{t},\ {y}={4}{t},\ {z}={8}-{8}{t}\)

Solving for t we have,

\(\displaystyle\Rightarrow{t}={\frac{{{x}+{1}}}{{{3}}}},\ {t}={\frac{{{y}}}{{{4}}}},\ {t}={\frac{{{z}-{8}}}{{-{8}}}}\)

\(\displaystyle\Rightarrow{\frac{{{x}+{1}}}{{{3}}}}={\frac{{{y}}}{{{4}}}}={\frac{{{z}-{8}}}{{-{8}}}}\)

\(\displaystyle\Rightarrow{\frac{{{x}+{1}}}{{{3}}}}={\frac{{{y}}}{{{4}}}}={\frac{{{8}-{z}}}{{{8}}}}\)

Symmetric equations

\(\displaystyle{\frac{{{x}+{1}}}{{{3}}}}={\frac{{{y}}}{{{4}}}}={\frac{{{8}-{z}}}{{{8}}}}\)

Given:

The given point is \(\displaystyle{\left(−{1},\ {0},\ {8}\right)}\) and the vector or line is \(\displaystyle{v}={3}{i}+{4}{j}-{8}{k}\)

To determine: The sets of parametric equations and symmetric equations of the line that passes through the given point and is parallel to the given vector or line.

Step 2

a) The parametric equations for a line passing through \(\displaystyle{\left({x}_{{{0}}},\ {y}_{{{0}}},\ {z}_{{{0}}}\right)}\) and parallel to the vector \(\displaystyle{v}={a}{i}+{b}{j}+{c}{k}\) are

\(\displaystyle{x}={x}_{{{0}}}+{a}{t},\ {y}={y}_{{{0}}}+{b}{t},\ {z}={z}_{{{0}}}+{c}{t}\)

The required parametric equations are

\(\displaystyle{x}=-{1}+{3}{t},\ {y}={4}{t},\ {z}={8}-{8}{t}\)

Step 3

b) The parametric equations are

\(\displaystyle{x}=-{1}+{3}{t},\ {y}={4}{t},\ {z}={8}-{8}{t}\)

Solving for t we have,

\(\displaystyle\Rightarrow{t}={\frac{{{x}+{1}}}{{{3}}}},\ {t}={\frac{{{y}}}{{{4}}}},\ {t}={\frac{{{z}-{8}}}{{-{8}}}}\)

\(\displaystyle\Rightarrow{\frac{{{x}+{1}}}{{{3}}}}={\frac{{{y}}}{{{4}}}}={\frac{{{z}-{8}}}{{-{8}}}}\)

\(\displaystyle\Rightarrow{\frac{{{x}+{1}}}{{{3}}}}={\frac{{{y}}}{{{4}}}}={\frac{{{8}-{z}}}{{{8}}}}\)

Symmetric equations

\(\displaystyle{\frac{{{x}+{1}}}{{{3}}}}={\frac{{{y}}}{{{4}}}}={\frac{{{8}-{z}}}{{{8}}}}\)