Question

Express the equations of the lines r · (i −j) + 7 = 0,r · (i + 3j) − 5 = 0 in parametric forms and hence find the position vector of their point of intersection.

Forms of linear equations
Express the equations of the lines
r · (i −j) + 7 = 0,r · (i + 3j) − 5 = 0 in parametric forms and hence find the position
vector of their point of intersection.

2020-11-03

Express the equation of the lines
$$\displaystyle{o}{v}{e}{r}\rightarrow{\left\lbrace{r}\right\rbrace}\cdot{\left({w}{i}{d}{e}\hat{{{i}}}-{w}{i}{d}{e}\hat{{{j}}}+{7}={0},{o}{v}{e}{r}\rightarrow{\left\lbrace{r}\right\rbrace}\cdot{\left({w}{i}{d}{e}\hat{{{i}}}+{3}{w}{i}{d}{e}\hat{{{j}}}\right)}-{5}={0}\right.}$$
in parametric form
$$\displaystyle{L}{e}{t}{w}{i}{d}{e}\hat{{{r}}}={x}{w}{i}{d}{e}\hat{{{i}}}+{y}{w}{i}{d}{e}\hat{{{j}}}+{z}{w}{i}{d}{e}\hat{{{k}}}$$
$$\displaystyle{w}{i}{d}{e}\hat{{{r}}}\cdot{\left({w}{i}{d}{e}\hat{{{i}}}-{w}{i}{d}{e}\hat{{{j}}}\right)}+{7}={0}$$
$$\displaystyle{\left({x}{w}{i}{d}{e}\hat{{{i}}}+{y}{w}{i}{d}{e}\hat{{{j}}}+{z}{w}{i}{d}{e}\hat{{{k}}}\right)}{\left({w}{i}{d}{e}\hat{{{i}}}{w}{i}{d}{e}\hat{{{j}}}\right)}+{7}={0}$$
$$\displaystyle{L}\in{e}{L}_{{1}}:-{x}-{y}+{7}={0}$$
$$\displaystyle{A}{n}{d}{w}{i}{d}{e}\hat{{{r}}}\cdot{\left({w}{i}{d}{e}\hat{{{i}}}+{3}{w}{i}{d}{e}\hat{{{j}}}\right)}-{5}={0}$$
$$\displaystyle{\left({x}{w}{i}{d}{e}\hat{{{i}}}+{y}{w}{i}{d}{e}\hat{{{j}}}+{z}{w}{i}{d}{e}\hat{{{k}}}\right)}{\left({w}{i}{d}{e}\hat{{{i}}}+{3}{w}{i}{d}{e}\hat{{{a}{b}}}\right)}-{5}={0}$$
$$\displaystyle{L}\in{e}{L}_{{2}}{x}+{3}{y}-{5}={0}$$
Line $$\displaystyle{L}_{{1}}$$ is in direction of $$\displaystyle{w}{i}{d}{e}\hat{{{i}}}-{w}{i}{d}{e}\hat{{{j}}}--{\left({v}\right)}$$
And $$\displaystyle{L}_{{1}}$$ also passes through (-3,4)(A)
Vector equation of $$\displaystyle{L}_{{1}}\Rightarrow{X}={A}+{t}{V}$$
$$\displaystyle{\left[{x},{y}\right]}={\left[-{3},{4}\right]}+{t}{\left[{1},{1}\right]}$$
$$\displaystyle{\left[{x},{y}\right]}={\left[-{3},{4}\right]}+{t}{\left[{t},{t}\right]}$$
$$\displaystyle{\left[{x},{y}\right]}={\left[-{3},+{t},{4}+{t}\right]}$$
in parametric form
$$\displaystyle{<}{x},{y}\ge{<}-{3}+{t},{4}+{t}{>}$$
Similarly,line $$\displaystyle{L}_{{2}}$$ also passes through (2,1)(A)
and direction vector [-3,1](v)
$$\displaystyle{L}_{{2}}={x}={a}+{t}{V}$$
$$\displaystyle{\left[{x},{y}\right]}={\left[{2},{1}\right]}+{t}_{{1}}{\left[-{3},{1}\right]}$$
$$\displaystyle{\left[{x},{y}\right]}={\left[{2}-{3}{t}_{{1}},{1}+{t}_{{1}}\right]}$$
$$\displaystyle{L}_{{2}}:\prec{x},{y}\ge{<}{2}-{3}{t}_{{1}},{1}+{t}_{{1}}{>}$$
Point of intersection of $$\displaystyle{L}_{{1}}{\quad\text{and}\quad}{L}_{{2}}$$ are:-
x=-4
So,Position vecor of their point of intersection
$$\displaystyle=-{4}{w}{i}{d}{e}\hat{{{i}}}+{3}{w}{i}{d}{e}\hat{{{j}}}$$