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Question
Find direction cosines of the normal to the plane `bar"r"*(3hat"i" + 4hat"k")` = 5
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Solution
Equation of the plane is `bar"r"*(3hat"i" + 4hat"k")` = 5
This is of the form,
`bar"r"*bar"n"` = 5, where `bar"n" = 3hat"i" + 4hat"k"`
Now, `|bar"n"| = sqrt(3^2 + 4^2)`
= `sqrt(9 + 16)`
= 5
The equation `bar"r"*bar"n"` = 5 can be written as
`bar"r"* (bar"n")/|bar"n"| = 5/|bar"n"|`
i.e., `bar"r"*(3/5hat"i" + 4/5hat"k") = 5/5`
= 1
∴ The direction cosines of the normal are `3/5, 0, 4/5`.
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The direction cosines of the line x - y + 2z = 5 and 3x + y + z = 6 are
