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प्रश्न
Answer the following question.
Show that vectors `vec"a" = 2hat"i" + 5hat"j" - 6hat"k" and vec"b" = hat"i" + 5/2 hat"j" - 3hat"k"` are parallel.
उत्तर
Let the angle between the two vectors be θ.
∴ cos θ = `(vec"a" * vec"b")/(|vec"a"||vec"b"|)`
`= ((2hat"i" + 5hat"j" - 6hat"k")*(hat"i" + 5/2hat"j" - 3hat"k"))/(sqrt(2^2 + 5^2 + (-6)^2)xxsqrt(1^2 + (5/2)^2 + (-3)^2))`
`= (2 + 25/2 + 18)/(sqrt65 xx sqrt(65//4))`
`= (65//2)/(65//2) = 1`
⇒ θ = cos-1 (1) = 0°
⇒ Two vectors are parallel.
Alternate method:
`vec"a" = 2hat"i" + 5hat"j" - 6hat"k" = 2 (hat"i" + 5/2hat"j" - 3hat"k") = 2 vec"b"`
Since `vec"a"` is a scalar multiple of `vec"b"`, the vectors are parallel.
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