If A= i-5j+7k and B= 3i+4j-6k then find out the angle between A and B, also calculate their vector product.
Answers
a= 2i+3j+4j
b=3i+4j+5k
The scalar product of these two vectors is-
a.b= |a| |b| cosθ
Where θ is the angle between the two vectors.
(2i+3j+4j)(3i+4j+5k) =
|a| |b| cosθ
Or,
6+12+20 = |a| |b| cosθ
Where,
|a| =sqrt(2^2+3^2+4^2)
|b|= sqrt(3^2+4^2+5^2)
Therefore
38= sqrt(29).sqrt(50)cosθ
cosθ= 38/sqrt(1450)
cosθ=38/38.07
cosθ= 0.998
θ= 1.54 degree.
The angle between the two vectors will be about 1.54 degree.
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Answer:
a= 2i+3j+4j
b=3i+4j+5k
The scalar product of these two vectors is
a.b= lal lbl cose
Where is the angle between the two
vectors.
(2i+3j+4j)(3i+4j+5k) =
lal lbl cose
Or,
6+12+20= lal lbl cose
Where,
lal =sqrt(2^2+3^2+4^2)
lbl= sqrt(3^2+4^2+5^2)
Therefore
38= sqrt(29).sqrt(50)cose
cose= 38/sqrt(1450)
cos0=38/38.07
cose= 0.998
0= 1.54 degree.
The angle between the two vectors will be about 1.54 degree.
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