Q .38. - Calculus

Question

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

$P(0,0,0), Q(1,2,3)$

Step-by-Step Solution

Answer

$\text{ The answer is }x\left(t\right)=t,y\left(t\right)=2t,z\left(t\right)=3t,0\le t\le 1\text{. }$

Step 1:Given information

$\text{ The points }P(0,0,0)\text{ and }Q(1,2,3)\text{. }$

Step 2:Calculation

$\text{ The point are }P(0,0,0)\text{ and }Q(1,2,3)\text{. }$

$\overrightarrow{PQ}=(1-0,2-0,3-0)$

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

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.

$\text{ Here }P(0,0,0)\text{ and }\overrightarrow{PQ}=d=(1,2,3)\text{ then the equation is, }$

$r\left(t\right)=(0,0,0)+t(1,2,3)$

$r\left(t\right)=(0+t, 0+2 t, 0+3 t)$

$\text{ The equation is }r\left(t\right)=(t,2t,3t)\text{. }$

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

$\text{ Then, }r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)=(t,2t,3t)$

$\text{ Thus the parametric equations are }x\left(t\right)=t,y\left(t\right)=2t,z\left(t\right)=3t\text{. }$

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 $x\left(t\right)=t, y\left(t\right)=2 t, z\left(t\right)=3 t$ where $0\le t\le 1$

$\text{ Therefore, the answer is }x\left(t\right)=t,y\left(t\right)=2t,z\left(t\right)=3t,0\le t\le 1\text{. }$

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