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Vector-A-of-magnitude-20-unit-lies-in-the-direction-45-S-of-E-while-vector-B-of-magnitude-30-units-in-the-direction-60-W-of-N-calculate-the-scaler-product-A-B-




Question Number 9715 by tawakalitu last updated on 28/Dec/16
Vector A^→  of magnitude 20 unit, lies in the   direction 45°S of E while vector B^→  of magnitude  30 units in the direction 60°W of N.  calculate the scaler product  A^→ ∙B^→
$$\mathrm{Vector}\:\overset{\rightarrow} {\mathrm{A}}\:\mathrm{of}\:\mathrm{magnitude}\:\mathrm{20}\:\mathrm{unit},\:\mathrm{lies}\:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{direction}\:\mathrm{45}°\mathrm{S}\:\mathrm{of}\:\mathrm{E}\:\mathrm{while}\:\mathrm{vector}\:\overset{\rightarrow} {\mathrm{B}}\:\mathrm{of}\:\mathrm{magnitude} \\ $$$$\mathrm{30}\:\mathrm{units}\:\mathrm{in}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{60}°\mathrm{W}\:\mathrm{of}\:\mathrm{N}. \\ $$$$\mathrm{calculate}\:\mathrm{the}\:\mathrm{scaler}\:\mathrm{product}\:\:\overset{\rightarrow} {\mathrm{A}}\centerdot\overset{\rightarrow} {\mathrm{B}} \\ $$
Answered by sandy_suhendra last updated on 28/Dec/16
Commented by sandy_suhendra last updated on 28/Dec/16
the angle between A^→  and B^→  = 165°  A^→ .B^→  = ∣A^→ ∣.∣B^→ ∣ cos 165°           = 20×30×(−0.966)           = −579.6 unit
$$\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\overset{\rightarrow} {\mathrm{A}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{B}}\:=\:\mathrm{165}° \\ $$$$\overset{\rightarrow} {\mathrm{A}}.\overset{\rightarrow} {\mathrm{B}}\:=\:\mid\overset{\rightarrow} {\mathrm{A}}\mid.\mid\overset{\rightarrow} {\mathrm{B}}\mid\:\mathrm{cos}\:\mathrm{165}° \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{20}×\mathrm{30}×\left(−\mathrm{0}.\mathrm{966}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:−\mathrm{579}.\mathrm{6}\:\mathrm{unit} \\ $$
Commented by tawakalitu last updated on 28/Dec/16
God bless you sir. i really appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Commented by geovane10math last updated on 29/Dec/16
Other way ...  60° + 45° + x = 270°  x = 270° − 105°  x = 165°
$${Other}\:{way}\:… \\ $$$$\mathrm{60}°\:+\:\mathrm{45}°\:+\:{x}\:=\:\mathrm{270}° \\ $$$${x}\:=\:\mathrm{270}°\:−\:\mathrm{105}° \\ $$$${x}\:=\:\mathrm{165}° \\ $$

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