Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ – plane.

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ – plane.

Solution:

It is known to us that the vector eq. of a line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is given as
$\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})$
Therefore the position vector of point $\mathrm{A}(5,1,6)$ is given as
$\vec{a}=5 \hat{i}+\hat{j}+6 \hat{k} \ldots$
And the position vector of point $\mathrm{B}(3,4,1)$ is given as $\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}} \ldots$
$\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}} \ldots(2)$
Therefore subtract eq.(2) and eq.(1) we obtain
$\begin{aligned} (\vec{b}-\vec{a}) =(3 \hat{i}+4 \hat{j}+\hat{k})-(5 \hat{i}+\hat{j}+6 \hat{k}) \\ =(3-5) \hat{i}+(4-1) \hat{j}+(1-6) \hat{k} \\ =(-2 \hat{i}+3 \hat{j}-5 \hat{k}) \\ \vec{r}=(5 \hat{i}+&\hat{j}+6 \hat{k})+\lambda(-2 \hat{i}+3 \hat{j}-5 \hat{k}) \\ \ldots .(3) \end{aligned}$

Let the coordinates of the point where the line crosses the YZ plane be $(0, \mathrm{y}, \mathrm{z})$ So,
$\vec{r}=(0 \hat{\boldsymbol{i}}+\boldsymbol{\boldsymbol { j }}+\boldsymbol{z} \hat{\boldsymbol{k}}) \ldots \ldots(4)$
As the point lies in line, it satisfies its eq.,

Substituting eq.(4) in eq.(3) we obtain,
$\begin{aligned} (0 \hat{\boldsymbol{i}}+\boldsymbol{y} \hat{\boldsymbol{j}}+\boldsymbol{z} \hat{\boldsymbol{k}}) &=(5 \hat{\boldsymbol{i}}+\hat{\boldsymbol{j}}+6 \hat{\boldsymbol{k}})+\lambda(-2 \hat{\boldsymbol{i}}+3 \hat{\boldsymbol{j}}-5 \hat{\boldsymbol{k}}) \\ =(5-2 \lambda) \hat{\boldsymbol{i}}+(1+3 \boldsymbol{\lambda}) \hat{\boldsymbol{j}}+(6-5 \lambda) \hat{\boldsymbol{k}} \end{aligned}$
It is known to us that, two vectors are equal if their corresponding components are equal

Therefore,
$0=5-2 \lambda$ $5=2 \lambda$ $\lambda=5 / 2$ $\mathrm{y}=1+3 \lambda$ And,
$z=6-5 \lambda \ldots(6)$

Substituting the value of $\lambda$ in eq. $(5)$ and $(6)$ , we obtain $-$
$\begin{array}{l} y=1+3 \lambda \\ =1+3 \times(5 / 2) \\ =1+(15 / 2) \end{array}$
$=17 / 2$