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Question
Show that vectors `vec"a" = 2hat"i" + 3hat"j" + 6hat"k", vec"b" = 3hat"i" - 6hat"j" + 2hat"k" and vec"c" = 6hat"i" + 2hat"j" - 3hat"k"` are mutually perpendicular.
Solution
As dot product of two perpendicular vectors is zero. Taking dot product of `vec"a" and vec"b"`
`vec"a"*vec"b" = (2hat"i" + 3hat"j" + 6hat"k")*(3hat"i" - 6hat"j" + 2hat"k")`
`= (2hat"i" xx 3hat"i")+(3hat"j" xx -6hat"j")+(6hat"k" xx 2hat"k")`
= 6 - 18 + 12
= 0
`vec"b"*vec"c" = (3hat"i" - 6hat"j" + 2hat"k")*(6hat"i" + 2hat"j" - 3hat"k")`
`= (3hat"i" xx 6hat"i")+(-6hat"j" xx 2hat"j")+(2hat"k" xx -3hat"k")`
= 18 - 12 - 6
= 0
Similarly,
`vec"c" * vec"a" = (6hat"i" + 2hat"j" - 3hat"k") * (2hat"i" + 3hat"j" + 6hat"k")`
= `(6hat"I" xx 2hat"I") + (2hat"j" xx 3hat"j") + (- 3hat"k" xx 6hat"k")`
= 12 + 6 - 18
= 0
As per the results, we can say that given three vectors `vec"a", vec"b" and vec"c"` are mutually perpendicular to each other.
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