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CBSE Class 12 Math 2020 Outside Delhi Set 2 Solved Paper

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Question : 8 of 11
Marks: +1, -0
Find the distance between the parallel planes
2x+y+2z=8 and 4x+2y+4z+5=02x+y+2z=8 \text{ and } 4x+2y+4z+5=0
Solution:     👈: Video Solution
We have
2x+y+27=82x+y+27=8
 and, 4x+2y+47=−5\text{ and, } 4x+2y+47=-5
⇒2x+y+27=−52\Rightarrow 2x+y+27=\frac{-5}{2}
Now,
 Distance d=∣8+52∣22+12+22\text{ Distance } d=\frac{\left|8+\frac{5}{2}\right|}{\sqrt{2^2+1^2+2^2}}
=2123=212×3=72=\frac{\frac{21}{2}}{\sqrt{3}}=\frac{21}{2\times 3}=\frac{7}{2}
Hence, the dis tan ce between two given lines are 72\frac{7}{2}.
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