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प्रश्न
Find the angle between the vectors `hati - 2hatj + 3hatk` and `3hati - 2hatj + hatk`.
बेरीज
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उत्तर
Let, `vec"a" = (hati - 2hatj + 3hatk)`
and `vec"b" = (3hati - 2hatj + hatk)`
Then `|veca| = sqrt(1^2 + (-2)^2 + 3^2)`
`= sqrt(1 + 4 + 9)`
`= sqrt14`
`|vecb| = sqrt(3^2 + (-2)^2 + 1^2)`
`= sqrt (9 + 4 + 1)`
`= sqrt14`
and `veca . vecb = (hati - 2hatj + 3hatk) * (3hati - 2hatj + hatk)`
= (1) (3) + (-2) (-2) + (3) (1)
= 3 + 4 + 3
= 10
If θ be the angle between `veca "and" vecb`, then
`cos theta = (veca . vec b)/(|veca| |vec b|)`
`= 10/(sqrt14 sqrt14)`
`= 10/14`
`= 5/7`
⇒ `theta = cos ^-1 5/7`
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