Question

The value of $\lambda$ for which the lines
$3x+2y+z-5=0=x+y-2z-3$ and
$2x-y-\lambda z=0=7x+10y-8z$
are perpendicular to each other is

Answer 1

Line$1:3x+2y+z-5=0=x+y-2z-3$

Line$2:2x-y-\lambda z=0=7x+10y-8z$

The required lines are Intersection of two planes.

$\therefore$ Let $l_1{m}_1{n}_1$ be directions of line $1$.

$\therefore 3l_1+2m_1+n_1=0$ (Since $l_1{m}_1{n}_1$ is perpendicular to the two planes)

$\therefore l_1+m_1-2n_1=0\therefore \frac{l_1}{-5}=\frac{-m_1}{-7}=\frac{n_1}{1}$

Let $l_2{m}_2{n}_2$ be directions of line $2$.

$2l_2-m_2-\lambda n_2=07l_2+10m_2-8n_2=0\frac{l_2}{8+10\lambda }=\frac{-m_2}{-16+7\lambda }=\frac{n_2}{27}$

$\left(l_1{m}_1{n}_1\right)$ and $\left(l_2{m}_2{n}_2\right)$ are directions of $\perp$ lines.

$\therefore l_1{l}_2+m_1{m}_2+n_1{n}_2=0-5(8+10\lambda )+7(16-7\lambda )+27=0\therefore \lambda =1$