Find the vector and the Cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).

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Find the vector and the Cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).

Solution:

It is given that
Let’s calculate the vector form:
The vector eq. of as line which passes through two points whose position vectors are $\vec{a}$ and $\vec{b}$ is $\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$
Here, the position vectors of the two points $(3,-2,-5)$ and $(3,-2,6)$ are $\vec{a}=3 \hat{i}-2 \hat{j}-5 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+6 \hat{k}$ respectively.
Therefore, the vector eq. of the required line is:
$begin{array}{l} \vec{r}=3 \hat{i}-2 \hat{j}-5 \hat{k}+\lambda[(3 \hat{i}-2 \hat{j}+6 \hat{k})-(3 \hat{i}-2 \hat{j}-5 \hat{k})] \\ \vec{r}=3 \hat{i}-2 \hat{j}-5 \hat{k}+\lambda(3 \hat{i}-2 \hat{j}+6 \hat{k}-3 \hat{i}+2 \hat{j}+5 \hat{k}) \\ \vec{r}=3 \hat{i}-2 \hat{j}-5 \hat{k}+\lambda(11 \hat{k}) \end{array}$
So now,
Using the formula,
The Cartesian equation of a line that passes through two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and $\left(\mathrm{x}_{2}\right.$ , $\left.\mathrm{y}_{2}, \mathrm{z}_{2}\right)$ is $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
Therefore, the Cartesian eq. of the line that passes through the origin $(3,-2,-5)$ and $(3,-2,6)$ is
$\frac{x-3}{3-3}=\frac{y-(-2)}{(-2)-(-2)}=\frac{z-(-5)}{6-(-5)}$ $\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}$ $\therefore$ The vector eq. is
$\overrightarrow{\mathrm{r}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}+\lambda(11 \hat{\mathrm{k}})$
The Cartesian eq. is
$\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}$