Q .6. - Calculus

Question

Let $P = (a, b,c)$and $Q = (\alpha , \beta , \gamma )$ be distinct points in ${\mathrm{ℝ}}^3$. Explain why the parametrization $x=a+(\alpha - a)t, y=b+(\beta - b)t, z=c+(\gamma - c)t$

Step-by-Step Solution

Answer

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Step 1:Given information

$\text{ Consider the distinct points }P=(a,b,c);Q=(\alpha ,\beta ,\gamma )\text{ in }{\mathrm{ℝ}}^3\text{. }$

Step 2:Explaination

First we will find the direction vector for the line $\overrightarrow{QP}$

 The points are $Q=(\alpha ,\beta ,\gamma )$ and $P=(a,b,c)$

$\overrightarrow{QP}=(a-\alpha ,b-\beta ,c-\gamma )$

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.

 Here$Q=(\alpha ,\beta ,\gamma )$ and  $\overrightarrow{QP}=d=(a-\alpha ,b-\beta ,c-\gamma )$ then the equation is,

$r\left(t\right)=(\alpha ,\beta ,\gamma )+t(a-\alpha ,b-\beta ,c-\gamma )$

$r\left(t\right)=(\alpha +(a-\alpha )t,\beta (b-\beta )t,\gamma +(c-\gamma \left)t\right)$

 The equation is written as follows, 

$r\left(t\right)=(\alpha +(a-\alpha )t,\beta (b-\beta )t,\gamma +(c-\gamma \left)t\right)$

Here the range is restricted so that the line segment starts and ends at the given points.

The line segment starts at $Q$ and ends at $P$

Thus $t$ is from $0$ to $1$ that is $0\le t\le 1$.

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

$r\left(t\right)=(\alpha +(a-\alpha )t,\beta (b-\beta )t,\gamma +(c-\gamma \left)t\right)\text{ where }0\le t\le 1$

$\text{ Therefore, the required equation is }r\left(t\right)=(\alpha +(a-\alpha )t,\beta (b-\beta )t,\gamma +(c-\gamma \left)t\right);0\le t\le 1\text{. }$