Question Vector 1 points along the z axis and has magnitude V1 = 80. Vector 2 lies in the xz plane, has magnitude V2 = 50, and makes a -37° angle with the x axis (points below the x axis). What is the scalar product V1·V2?

Answer: [tex]\vec{V1}[/tex] . [tex]\vec{V2}[/tex] = 2574.08 Explanation: given data magnitude V1 = 80 magnitude V2 = 50 angle a = -37° solution [tex]\vec{V1}[/tex] = 80 k [tex]\vec{V2}[/tex] = 50 cos{37} i – 50sin{37} k so that here [tex]\vec{V1}[/tex] . [tex]\vec{V2}[/tex] is [tex]\vec{V1}[/tex] . [tex]\vec{V2}[/tex] = 80 k . ( 50 cos{37} i – 50sin{37} k ) [tex]\vec{V1}[/tex] . [tex]\vec{V2}[/tex] = 80 k . ( 38.270 i + 32.176 k ) [tex]\vec{V1}[/tex] . [tex]\vec{V2}[/tex] = 2574.08 Log in to Reply

Answer:[tex]\vec{V1}[/tex]

.[tex]\vec{V2}[/tex]= 2574.08Explanation:given data

magnitude V1 = 80

magnitude V2 = 50

angle a = -37°

solution

[tex]\vec{V1}[/tex] = 80 k

[tex]\vec{V2}[/tex] = 50 cos{37} i – 50sin{37} k

so that here [tex]\vec{V1}[/tex] . [tex]\vec{V2}[/tex] is

[tex]\vec{V1}[/tex] . [tex]\vec{V2}[/tex] = 80 k . ( 50 cos{37} i – 50sin{37} k )

[tex]\vec{V1}[/tex] . [tex]\vec{V2}[/tex] = 80 k . ( 38.270 i + 32.176 k )

[tex]\vec{V1}[/tex]

.[tex]\vec{V2}[/tex]= 2574.08