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Question
Show that each of the given three vectors is a unit vector:
`1/7 (2hati + 3hatj + 6hatj), 1/7(3hati - 6hatj + 2hatk), 1/7(6hati + 2hatj - 3hatk)`
Also, show that they are mutually perpendicular to each other.
Solution
Let `veca = 1/7 (2hati + 3hatj + 6hatk), vecb = 1/7 (3hati - 6hatj + 2hatk) "and" vecc = 1/7 (6hati + 2hatj - 3hatk)`
∴ `|veca| = sqrt((2/7)^2 + (3/7)^2 + (6/7)^2)`
`= sqrt(4/49 + 9/49 + 36/49)`
`= sqrt(49/49)`
`= sqrt1`
= 1
`|vecb| = sqrt((3/7)^2 + ((-6)/7)^2 + (2/7)^2)`
`= sqrt(9/49 + 36/49 + 4/49)`
`= sqrt(49/49)`
`= sqrt1`
= 1
`|vecc| = sqrt((6/7)^2 + (2/7)^2 + ((-3)/7)^2)`
`= sqrt(49/49)`
`= sqrt1`
= 1
Hence `vec a, vec b, vecc` are unit vectors.
Now, `veca . vecb = 1/49 [(2) (3) + (3) (-6) + (6) (2)]`
`= 1/49 [6 - 18 + 12]`
= 0
So, `veca` is perpendicular to `vecb`
`vecb . vecc = 1/49 [(3). (6) + (-6) (2) + (2) (-3)]`
`= 1/49 [18 - 12 - 6]`
= 0
So, `vecb` is perpendicular to `vecc.`
`vecc . vec a = 1/49 [(6) (2) + (2) (3) + (-3) (6)]`
`= 1/49 [12 + 6 - 18]`
= 0
So, `vecc` is perpendicular to `vec a`
Hence, `vec a, vecb "and" vec c` are three mutually perpendicular unit vectors.
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