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

The Correct Option is D

\(Δ=\begin{vmatrix} 8 & 1 & 4 \\[0.3em] 1 & 1 & 1 \\[0.3em] λ & -3 & 0 \end{vmatrix}\)
\(=8(3)−1(−λ)+4(−3−λ)\)
\(=24+λ–12–4λ\)
\(=12–3λ\)
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): \(\frac {10}{3}\)