Q .37. - Calculus

Question

use the result of Exercise 35 to find parametric equations for the line segment connecting point P to point Q. 

$P(1,7,3), Q(-1,-2,5)$

Step-by-Step Solution

Answer

$\text{ The answer is }x\left(t\right)=1-2t,y\left(t\right)=7-9t,z\left(t\right)=3+2t\text{. }$

Step 1:Given information

$\text{The points }P(1,7,3)\text{ and }Q(-1,-2,5)\text{. }$

Step 2: Calculation

$\text{ The point are }P(1,7,3)\text{ and }Q(-1,-2,5)\text{. }$

$\overrightarrow{PQ}=(-1-1,-2-7,5-3)$

$\overrightarrow{PQ}=(-2,-9,2)$

The formula to find the line $\mathcal{L}$ equation is as follows, $r\left(t\right)=P_0+td$ Where, $P_0$ is the point and d is the direction vector.

$\text{ Here }P(1,7,3)\text{ and }\overrightarrow{PQ}=d=(-2,-9,2)\text{ then the equation is, }$

$r\left(t\right)=(1,7,3)+t(-2,-9,2)$

$r\left(t\right)=(1-2 t, 7-9 t, 3+2 t)$

$\text{ The equation is }r\left(t\right)=(1-2t,7-9t,3+2t)\text{. }$

$\text{ The vector function }r\left(t\right)\text{ in three -dimensional plane represents }r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)\text{. }$

$\text{ Then, }r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)=(1-2t,7-9t,3+2t)$

$\text{ The parametric equations are }x\left(t\right)=1-2t,y\left(t\right)=7-9t,z\left(t\right)=3+2t$

The restriction for the parameter $t$ so that the result parametrizes the segment P to point Q is given from 0 to 1 that is from $0\le t\le 1$

Thus, the parametric equations are $x\left(t\right)=1-2 t, y\left(t\right)=7-9 t, z\left(t\right)=3+2 t$ where $0\le t\le 1$

$\text{ Therefore, the answer is }x\left(t\right)=1-2t,y\left(t\right)=7-9t,z\left(t\right)=3+2t\text{. }$