Question

. Find an equation of the line containing the point $(x_0,y_0,z_0)$ with direction vector (a, b,c).

(a) Use a different direction vector to find another equation for the same line. (b) Assuming that a = 0, find an equation for the same line with the form

$x=at+5,y=bt+y_0,z=ct+z_0$

Short answer

Part a)$\text{The required equation is}r\left(t\right)=\left(x_0+2at,y_0+2bt,z_0+2ct\right)\text{.}$

Part b)The required equation is$r\left(t\right)=\left(x_0+5t,y_0+t\left(y_0+\frac{b\left(5-x_0\right)}{a}\right),z_0+t\left(z_0+\frac{c\left(5-x_0\right)}{a}\right)\right)$

Step-by-step solution

Part (a) Step 1:Given information

$\text{Consider the point}\left(x_0,y_0,z_0\right)\text{and the direction vector}(a,b,c)\text{.}$

 part (b) Step 2:Calculation

$\text{Consider the point}\left(x_0,y_0,z_0\right)\text{and the direction vector}(a,b,c)\text{.}$

$\text{The formula to find the line}\mathcal{L}\text{equation is as follows,}$

$\text{For}P\left(x_0,y_0,z_0\right),d=(a,b,c)\text{the equation is,}$

$r\left(t\right)=\left(x_0,y_0,z_0\right)+t(a,b,c)$

The equation is written as follows,

$r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)$

$\text{The equation}\mathcal{L}\text{in the form of vector parametrization is}r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)\text{.}$

$\text{Therefore, the required equation is}r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)\text{.}$

The direction vector is$(a,b,c).$

Objective is to find another equation for the same line using different direction vectors.

Another direction vector is be taken by multiplying the direction vector by a scalar.

Thus, the different direction vector is $d=(a,b,c)=2(a,b,c).$

$d=(2a,2b,2c)$

$\text{Now for}P\left(x_0,y_0,z_0\right),d=(2a,2b,2c)\text{the equation is,}$

$r\left(t\right)=\left(x_0,y_0,z_0\right)+t(2a,2b,2c)$

$r\left(t\right)=\left(x_0+2at,y_0+2bt,z_0+2ct\right)$

$\text{The equation}\mathcal{L}\text{is}r\left(t\right)=\left(x_0+2at,y_0+2bt,z_0+2ct\right)$

$\text{Therefore, the required equation is}r\left(t\right)=\left(x_0+2at,y_0+2bt,z_0+2ct\right)\text{.}$

Part (b) Step 1:Given information

$\text{Consider the equations}x=at+5,y=bt+y_0,z=ct+z_0\text{.}$

Part (b) Step 2:Calculation

$\text{Consider the equations}x=at+5,y=bt+y_0,z=ct+z_0\text{.}$

$\text{From the equations}x=at+5,y=bt+y_0,z=ct+z_0\text{the point is}\left(5,y_0,z_0\right)\text{.}$

Now equate the value of $x$to the value given in the equation.

$r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)$and find the value of $t$.

By equating them,

$x_0+at=5$

Add $x_0$on both sides of the equation.

$x_0+at-x_0=5-x_0$

$\Rightarrow t=5-x_0$

$\text{Now substitute}t=5-x_0\text{in the equation}r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)\text{.}$

Then,

$\left(x_0+at,y_0+bt,z_0+ct\right)=\left(5,y_0+b\frac{5-x_0}{a},z_0+c\frac{5-x_0}{a}\right)$

$=\left(5,y_0+\frac{b\left(5-x_0\right)}{a},z_0+\frac{c\left(5-x_0\right)}{a}\right)$

$\text{The line equation for the point}\left(5,y_0+\frac{b\left(5-x_0\right)}{a},z_0+\frac{c\left(5-x_0\right)}{a}\right)\text{is,}$

$r\left(t\right)=\left(x_0,y_0,z_0\right)+t\left(5,y_0+\frac{b\left(5-x_0\right)}{a},z_0+\frac{c\left(5-x_0\right)}{a}\right)$

The equation is written as follows,

$r\left(t\right)=\left(x_0+5t,y_0+t\left(y_0+\frac{b\left(5-x_0\right)}{a}\right),z_0+t\left(z_0+\frac{c\left(5-x_0\right)}{a}\right)\right)$

The equation $\mathcal{L}$ in the form of vector parametrization is

$r\left(t\right)=\left(x_0+5t,y_0+t\left(y_0+\frac{b\left(5-x_0\right)}{a}\right),z_0+t\left(z_0+\frac{c\left(5-x_0\right)}{a}\right)\right)$

Therefore, the required equation is

r$\left(t\right)=\left(x_0+5t,y_0+t\left(y_0+\frac{b\left(5-x_0\right)}{a}\right),z_0+t\left(z_0+\frac{c\left(5-x_0\right)}{a}\right)\right)$