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(I)(a) Find the scalar product of the vectorsA→ andB→given in Exercise 1.38. (b) Find the angle between these two vectors.

Short Answer

Expert verified

The angle between these two vector is 1.43°.

Step by step solution

01

Given data

A→=4.00i^+7.00j^B→=5.00i^-2.00j^

02

The scalar product of vector

a.b=4.00i^+7.00j^.5.00i^-7200j^=4×5+7×-2=20-14=6

03

The angle between two vector

The magnitude of vector is,

A→=42+72=65=8.06B→=52+22=29=5.3

The angle between the two vector is,

θ=cos-1a→.b→a→b→=cos-168.065.3=cos-1642.87=cos-10.1399θ=1.43°

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