Question

Prove that lines $\frac{x-3}{3}=\frac{y-3}{-4}=\frac{z-5}{2}$ and $\frac{x}{6}=\frac{y-5}{-8}=\frac{z-2}{4}$ are parallel. Also find the equation of plane passing through them.

Answer 1

Given line L$:\frac{x-3}{3}=\frac{y-3}{-4}=\frac{z-5}{2}$
And $M:\frac{x}{6}=\frac{y-5}{-8}=\frac{z-3}{4}$
Direction of first line $\overline{l}=(3,-4,2)$
Direction of second line $\overline{m}=(6,-8,4)$
Now $\overline{l}\times \overline{m}=\left|\begin{array}{ccc}i& j& k\\ 3& -4& 2\\ 6& -8& 2\end{array}\right|$
$=i(-16+16)-j(12-12)+k(-24+24)$
$=(0,0,0)$
$=\overline{0}$
$\overline{l}\times \overline{m}=\overline{0}$
$\therefore$ line l $=$ line m OR line L $||$ line M
$\therefore$ From first equation of line we have
$A\left(\overline{a}\right)(3,3,5)$
$\therefore$ Line $M:\frac{x}{6}=\frac{y-5}{-8}=\frac{z-3}{4}$
$\therefore \frac{3}{6}=\frac{3-5}{-8}=\frac{5-2}{4}$ which is not valid
$\therefore (3,3,5)\text{/}\in M$
$\therefore L\ne M$
$\therefore L||M$
$\therefore$ Given lines are parallel equation of the plane passes from parallel lines in as follows
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ l_1& l_2& l_3\end{array}\right|=0$
$\therefore$ Required plane, $\left|\begin{array}{ccc}x-3& y-3& z-5\\ 0-3& 5-3& 2-5\\ 3& -4& 2\end{array}\right|=0$
$\therefore \left|\begin{array}{ccc}x-3& y-3& z-5\\ -3& 2& -3\\ 3& -4& 2\end{array}\right|=0$
$\therefore (x-3)(4-12)-(y-3)$
$(-6+9)+(z-5)(12-6)=0$
$\therefore (x-3)(-8)-(y-3)\left(3\right)+(z-5)6=0$
$\therefore -8x+24-3y+9+6z-30=0$
$\therefore -8y-3y+6z+3=0$
$\therefore 8x+3y-6z=3$
Which is required equation of plane.