If the system of linear equations.

Question

$8x + y + 4z =  - 2$

$x + y + z = 0$

$$\lambda x - 3y = \mu $$

has infinitely many solutions, then the distance of the point $$\left( {\lambda ,\mu , - {1 \over 2}} \right)$$ from the plane $8x + y + 4z + 2 = 0$ is :

Accepted Answer

$$\Delta  = \left| {\begin{pmatrix} 8 & 1 & 4  \
   1 & 1 & 1  \
   \lambda  & { - 3} & 0  \

\end{pmatrix} } \right|$$

$$ = 8(3) - 1( - \lambda ) + 4( - 3 - \lambda )$$

$$ = 24 + \lambda  - 12 - 4\lambda $$

$$ = 12 - 3\lambda $$

So for $$\lambda  = 4$$, it is having infinitely many solutions.

$${\Delta _x} = \left| {\begin{pmatrix} { - 2} & 1 & 4  \
   0 & 1 & 1  \
   \mu  & { - 3} & 0  \

\end{pmatrix} } \right|$$

$$ =  - 2(3) - 1( - \mu )+4( - \mu )$$

$$ =  - 6 - 3\mu  = 0$$

For $$\mu  =  - 2$$

Distance of $$(4, - 2,{{ - 1} \over 2})$$ from $8x + y + 4z + 2 = 0$

$$ = {{32 - 2 - 2 + 2} \over {\sqrt {64 + 1 + 16} }} = {{10} \over 3}$$ units

Solutions · 1