Question

If the system of linear equations.
$8x + y \+ 4z = –2$
$x + y + z = 0$
$λx–3y=μ$
has infinitely many solutions, then the distance of the point $(λ, μ, -1/2)$ from the plane $8x + y \+ 4z \+ 2 = 0$ is

• A. $3\sqrt 5$
• B. $4$
• C. $\frac {16}{9}$
• D. $\frac {10}{3}$

Answer 1

Answer: (D)

$\frac {10}{3}$

Answer 2

$Δ=\begin{vmatrix} 8 & 1 & 4 \\\[0.3em] 1 & 1 & 1 \\\[0.3em] λ & -3 & 0 \end{vmatrix}$
$=8(3)−1(−λ)+4(−3−λ)$
$=24+λ–12–4λ$
$=12–3λ$
So for $λ = 4$, it is having infinitely many solutions.
$Δ_x= \begin{vmatrix} -2 & 1 & 4 \\\[0.3em] 0 & 1 & 1 \\\[0.3em] μ & -3 & 0 \end{vmatrix}$
$Δ_x=−2(3)−1(−μ)+4(−μ)$
$Δ_x=−6–3μ$
$Δ_x=0$
For $μ= –2$
Distance of $(4,−2,−\frac 12)$ from $8x+y+4z+2=0$
$=\frac {32−2−2+2}{\sqrt {64+1+16}}$

$=\frac {10}{3}$ units

So, the correct option is (D): $\frac {10}{3}$