The line through the point

klytamnestra9a
2021-11-11
Answered

Find a vector equation and parametric equations for the line.

The line through the point$(1,0,8)$ and perpendicular to the plane $x+3y+z=3$

The line through the point

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Nancy Johnson

Answered 2021-11-12
Author has **17** answers

The line passing through the point (1,0,8) and perpendicular to the plane $x+3y+z=3$

Let${r}_{0}=i+0j+8k$

A line perpendicular to the plane$x+3y+z=3$ has the same direction as a normal vector to the plane, which is $n=<1,3,1>$

So, take$v=i+3j+k$

By using the vector eqaution for the line$r={r}_{0}+tv$

$r=(i+0j+8k)+t(i+3j+k)$

$=(1+t)i+(0+3t)j+(8+t)k$

$=(1+t)i\left(3t\right)j+(8+t)k$

Thus, the vector equation of the line is$r\left(t\right)=(1+t)+i\left(3t\right)j+(8+t)k$

Obtain the parametric equation of line as follows.

Let${P}_{0}=({x}_{0},{y}_{0},{z}_{0})=(1,0,8)$ and $v=<a,b,c\ge <1,3,1>$

By the parametric equation for the line through the point$({x}_{0},{y}_{0},{z}_{0})$ and parallel to the direction vector $<a,b,c>$ are $x={x}_{0}+at,\text{}y={y}_{0}+bt,\text{}z={z}_{0}+ct$

That implies,

$x=1+t,\text{}y=0+3t,\text{}z=8+t$

$x=1+t,\text{}y=3t,z=8+t$

Thus, the parametric equation of the line is

$x\left(t\right)=1+t,\text{}y\left(t\right)=3t,\text{}z\left(t\right)=8+t$

Let

A line perpendicular to the plane

So, take

By using the vector eqaution for the line

Thus, the vector equation of the line is

Obtain the parametric equation of line as follows.

Let

By the parametric equation for the line through the point

That implies,

Thus, the parametric equation of the line is

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