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the angle between the planes r.(3i-4j+5k)=0 & r.(2i-j-2k) is

the angle between the planes r.(3i-4j+5k)=0 & r.(2i-j-2k) is

Grade:Upto college level

1 Answers

Latika Leekha
askIITians Faculty 165 Points
7 years ago
The given planes are (3i-4j+5k)=0 & (2i-j-2k) .
Let θ be the angle between two planes, then we have
cos θ = (l1.l2) / |l1| . |l2 |, where l1 and l2 are the two vectors
Here, l1.l2 = 3(2) + (-4)(-1) + (5)(-2)
= 6+4-10
= 0.
Now, |l1 | = √ 32 + 42+ 52 = √ 50
and |l2 | = √ 22 + 12+ 22 = √ 9 = 3.
Hence cos θ = (l1.l2) / |l1| . |l2 |
= 0/ √ 50 .3
= 0.
Now cos θ = 0 which means θ = π/2.
Thanks & Regards
Latika Leekha
askIITians Faculty

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