JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the system of linear equations . \(8 x+y+4 z=-2\) ; \(x+y+z=0\) ; \(\lambda x-3 y=\mu\) has infinitely many solutions, then the distance of the point \(\left(\lambda, \mu,-\frac{1}{2}\right)\) from the plane \(8 x+y+4 z+\) \(2=0\) is.
- A \(3 \sqrt{5}\)
- B \(4\)
- C \(\frac{26}{9}\)
- D \(\frac{10}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{10}{3}\)
Step-by-step Solution
Detailed explanation
\(D =\left|\begin{array}{ccc}8 & 1 & 4 \\1 & 1 & 1 \\\lambda & -3 & 0\end{array}\right|=0 \Rightarrow \lambda=4\) Also \(D _{1}= D _{2}= D _{3}=0\) So \( \mu=-2\) Point \(\left(4,-2,-\frac{1}{2}\right)\) Distance from plane \(=\frac{10}{3}\)
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