Problem

Find the angle between vectors u→=(3,4,0) and v→=(2,1,−2) using the cross product.

Answer

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Answer

The angle θ between the vectors can be found using the formula sin⁡θ=‖u→×v→‖‖u→‖‖v→‖. Substituting the computed values, we get sin⁡θ=895∗3=8915. Thus, θ=arcsin⁡(8915).

Steps

Step 1 ：First, calculate the cross product of u→ and v→. The cross product of u→ and v→ is given by u→×v→=(u2v3−u3v2,u3v1−u1v3,u1v2−u2v1). Substituting the given values, we get u→×v→=(4∗(−2)−0∗1,0∗2−3∗(−2),3∗1−4∗2)=(−8,6,−5).

Step 2 ：The magnitude of a vector x→=(x1,x2,x3) is given by ‖x→‖=x12+x22+x32. Applying this, we get ‖u→‖=32+42+02=5 and ‖v→‖=22+12+(−2)2=3. The magnitude of the cross product ‖u→×v→‖=(−8)2+62+(−5)2=89.

Step 3 ：The angle θ between the vectors can be found using the formula sin⁡θ=‖u→×v→‖‖u→‖‖v→‖. Substituting the computed values, we get sin⁡θ=895∗3=8915. Thus, θ=arcsin⁡(8915).

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