AP EAMCET · Maths · Three Dimensional Geometry
If the planes \(2 x+3 y+4 z+7=0\) and \(4 x+k y+8 z+1\) \(=0\) are parallel, then the equation of the plane passing through is point \((\mathrm{k}, \mathrm{k}, \mathrm{k})\) and having the direction ratios of its normal as \((\mathrm{k}-1, \mathrm{k}, \mathrm{k}+1)\) is
- A \(x+2 y+3 z=36\)
- B \(3 x+4 y+5 z=72\)
- C \(4 x+5 y+6 z=90\)
- D \(5 x+6 y+7 z=108\)
Answer & Solution
Correct Answer
(D) \(5 x+6 y+7 z=108\)
Step-by-step Solution
Detailed explanation
Since given planes, \(2 x+3 y+4 z+7=0\) and \(4 x+k y+8 z+1=0\) are parallel Hence \(\frac{2}{4}=\frac{3}{k} \Rightarrow k=6\) Hence now the required plane through point \((k, k, k)\) and having the direction ratios of its normal as \((k-1, k, k+1)\) is…
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