Find sets of parametric equations and symmetric equations of the

embaseclielenzn 2021-11-26 Answered
Find 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.
\[\begin{array}{|c|c|}\hline \text{Point} & \text{Parallel to}v= \\ \hline (0,\ 0,\ 0) & <<8,\ 1,\ 4>> \\ \hline \end{array}\]
(a) parametric equations
(b) symmetric equations
\(\displaystyle{8}{x}={y}={4}{z}\)
\(\displaystyle{4}{x}={y}={8}{z}\)
\(\displaystyle{\frac{{{x}}}{{{8}}}}={y}={\frac{{{z}}}{{{4}}}}\)
\(\displaystyle{\frac{{{x}}}{{{4}}}}={y}={\frac{{{z}}}{{{8}}}}\)

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Expert Answer

Richard Cheatham
Answered 2021-11-27 Author has 6611 answers
Step 1
consider the t equation
Point \(\displaystyle={\left({0},\ {0},\ {0}\right)}\)
\(\displaystyle{v}={\left({8},\ {1},\ {4}\right)}\)
Step 2
Sets of parametric equations and symmetric equations of the line that passes through the given point
Point \(\displaystyle={\left({0},\ {0},\ {0}\right)}\)
\(\displaystyle{v}={\left({8},\ {1},\ {4}\right)}\)
To find a set of parametric equations of the line, we use the coordinates:
\(\displaystyle{x}_{{{1}}}={0}\)
\(\displaystyle{y}_{{{1}}}={0}\)
\(\displaystyle{z}_{{{1}}}={0}\)
The direction numbers:
\(\displaystyle{a}={8},\ {b}={1},\ {c}={4}\)
The parametric equations are:
\(\displaystyle{x}={0}+{8}={8}\)
\(\displaystyle{y}={0}+{1}={1}\)
\(\displaystyle{z}={0}+{4}={4}\)
(b) symmetric equation for
\(\displaystyle{8}{x}={y}={4}{z}\)
\(\displaystyle{\frac{{{x}-{x}_{{{1}}}}}{{{a}}}}={\frac{{{y}-{y}_{{{1}}}}}{{{b}}}}={\frac{{{z}-{z}_{{{1}}}}}{{{c}}}}\)
Substitute \(\displaystyle{\left({x},\ {y},\ {z}\right)}={\left({0},\ {0},\ {0}\right)}{n}={\left\langle{a},\ {b},\ {c}\right\rangle}={\left\langle{8},\ {1},\ {4}\right\rangle}\)
in symmetric equation
\(\displaystyle{\frac{{{x}-{0}}}{{{8}}}}={\frac{{{y}-{0}}}{{{1}}}}={\frac{{{z}-{3}}}{{{4}}}}\)
\(\displaystyle{\frac{{{8}-{8}}}{{{0}}}}={\frac{{{y}-{0}}}{{{1}}}}={\frac{{{4}-{3}}}{{{4}}}}\)
\(\displaystyle{\frac{{{8}}}{{{8}}}}={\frac{{{y}-{0}}}{{{1}}}}={\frac{{{1}}}{{{4}}}}\)
Direction number as integers \(\displaystyle{1},\ {1},\ {4}.\)
symmetric equation for \(\displaystyle{4}{x}={y}-{8}{Z}\)
substitute \(\displaystyle{\left({x},\ {y},\ {z}\right)}={\left({0},\ {0},\ {0}\right)}{n}={\left\langle{a},\ {b},\ {c}\right\rangle}={\left\langle{8},\ {1},\ {4}\right\rangle}{\frac{{{x}-{0}}}{{{8}}}}={\frac{{{y}-{0}}}{{{1}}}}={\frac{{{z}-{3}}}{{{4}}}}\)
in symmetric equation \(\displaystyle{\frac{{{0}-{8}}}{{{8}}}}={\frac{{{0}-{0}}}{{{1}}}}={\frac{{{4}-{3}}}{{{4}}}}\)
\(\displaystyle{\frac{{{4}-{0}}}{{{8}}}}={\frac{{{0}-{0}}}{{{1}}}}={\frac{{{8}-{4}}}{{{4}}}}\) direction number as integer 1, 1, 4
\(\displaystyle{\frac{{{4}}}{{{8}}}}={\frac{{{0}}}{{{1}}}}={\frac{{{4}}}{{{4}}}}\)
Direction number as integer 2, 1, 1
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