JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the system of linear equations \(-x+2 y-9 z=7\) \(-x+3 y+7 z=9\) \(-2 x+y+5 z=8\) \(-3 x+y+13 z=\lambda\) has a unique solution \(x =\alpha, y =\beta, z =\gamma\). Then the distance of the point \((\alpha, \beta, \gamma)\) from the plane \(2 x-2 y+z=\lambda\) is
- A \(9\)
- B \(11\)
- C \(13\)
- D \(7\)
Answer & Solution
Correct Answer
(D) \(7\)
Step-by-step Solution
Detailed explanation
\(-x+2 y-9 z=7-(1)\) \(-x+3 y-7 z=9-(2)\) \(-2 x+y+5 z=8-(3)\) \((2)-(1)\) \(y+16 z=2\) \((3)-2 \times(1)\) \(-3 y+23 z=-6-(5)\) \(3 \times(4)+(5)\) \(71 z=0 \Rightarrow z=0\) \(\quad y=2\) \(\quad x=-3\) \((-3,2,0) \rightarrow(\alpha, \beta, \gamma)\) Put in…
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