Q .10. - Calculus

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 

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

Step-by-Step Solution

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

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

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

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

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

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

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

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

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