Question

Prove that the lines
$\left[\frac{x+3}{3}=\frac{y+3}{5}=\frac{z+5}{7}\right]\&\left[\frac{x+2}{1}=\frac{y-4}{3}=\frac{z-6}{5}\right]$ not intersect at the point $\left[\frac{1}{2},\frac{-1}{2},\frac{-3}{2}\right]$?

Answer 1

Lines are as

$\frac{x+3}{3}=\frac{y+3}{5}=\frac{z+5}{7}={\lambda }_1$

$\frac{x+2}{1}=\frac{y-4}{3}=\frac{z-6}{5}={\lambda }_2$

From equation (i) we can write .

$x_1=3{\lambda }_1-3$;

$y_1=5{\lambda }_1-3$;

$z_1=7{\lambda }_1-5$

From equation (ii) we can write

$x_1={\lambda }_2-2$;

$y_1=3{\lambda }_2+4$;

$z_1=5{\lambda }_2+6$

So, $x_1=3{\lambda }_1-3={\lambda }_2-2$

$3{\lambda }_1-{\lambda }_2=1$ ...(iii)

Similarly,

$y_1=6{\lambda }_1-3=3{\lambda }_2+4$

$5{\lambda }_1-3{\lambda }_2=7$ ...(iv)

${\lambda }_2=3{\lambda }_1-1$ substitute in equation ......(iv)

$\Rightarrow 5{\lambda }_1-3(3{\lambda }_1-1)=7$

$\Rightarrow 5{\lambda }_1-9{\lambda }_1+3=7$

$\Rightarrow -4{\lambda }_1=4$

$\Rightarrow {\lambda }_1=-1$

${\lambda }_2=3{\lambda }_1-1=3(-1)-1=-4$

$\Rightarrow -4$

$\Rightarrow {\lambda }_2=-4$

Now if we check $z_1=7{\lambda }_1-5=5{\lambda }_2+6$

$7(-1)-5=5(-4)+6$

$-12=-14$ not possible

So, Both lines will not interset each other