Question

Find the co ordinates of the point where the line through $(3,4,1)$ and $(5,1,6)$ crosses $XY-$plane.

Answer 1

The equation of the line passing through the two points with vector

$\overrightarrow{a}+\overrightarrow{b}$

$\overrightarrow{r}=\overrightarrow{a}+\lambda (\overrightarrow{b}-\overrightarrow{a})$

since the line passes through the point is:

$\overrightarrow{a}=3\hat{i}+4\hat{j}+\hat{k}$ and $\overrightarrow{a}=5\hat{i}+\hat{j}+6\hat{k}$

$(\overrightarrow{b}-\overrightarrow{a})=5\hat{i}+\hat{j}+6\hat{k}-\overrightarrow{a}=3\hat{i}+4\hat{j}+\hat{k}$

$=2\hat{i}-3\hat{j}+5\hat{k}$

$\overrightarrow{r}=3\hat{i}+4\hat{j}+\hat{k}+\lambda (2\hat{i}-3\hat{j}+5\hat{k}).....1$

Now,

The coordinates of the point where the line cross the XY plane by (X,Y,0)

so,

$\overrightarrow{r}=x\hat{i}+y\hat{j}+0\hat{k}....2$

Since, these are crossing the points so put this value in equation 1 then we get,

$x\hat{i}+y\hat{j}+0\hat{k}=3\hat{i}+4\hat{j}+\hat{k}+\lambda (2\hat{i}-3\hat{j}+5\hat{k})$

Equating the coefficient of the unit vector then,

$x=1+5\lambda ...a$

$y=4-3\lambda ...b$

$1+5\lambda =\mathrm{0...}c$

by solving equation $c$ we get

$1+5\lambda$

$\lambda =-\frac{1}{5}$

This is the required solution.