Vector A with arrow has a magnitude of 64 units and points due west, while vector B with arrow has the same magnitude and points due south. Specify the direction relative to due west. Find the magnitude and direction of A with arrow and B with arrow.

Answer:The magnitude is 90.5.The direction of  [tex](\vec{A}+ \vec{B})[/tex] is 45° south of west.The direction of  [tex](\vec{A} - \vec{B})[/tex] is 45° north of west.Step-by-step explanation:Let suppose the direction of unit vector as follows:[tex]\hat{i}[/tex] in the eastern direction[tex]\hat{j}[/tex] in the northern directionGiven:Vector A with arrow has a magnitude of 64 units and points due west, while vector B with arrow has the same magnitude and points due south.Therefore,[tex]\vec{A} = -64 \hat{i}[/tex][tex]|\vec{A}| = 64[/tex]And[tex]\vec{B} = -64 \hat{j}[/tex][tex]|\vec{B}| = 64[/tex]Formula used:[tex]The\ magnitude\ of\ vector\ (\vec{A}+ \vec{B}) = \sqrt{|A|^{2}+|B|^{2}}[/tex]And the direction is given by:[tex]tan\theta = \frac{|A|}{|B|}[/tex]Now,The magnitude of [tex](\vec{A}+ \vec{B})[/tex]: [tex]= \sqrt{|A|^{2}+|B|^{2}}[/tex] [tex]= \sqrt{64^{2}+64^{2}}[/tex] [tex]= 64\sqrt{2}[/tex] [tex]= 90.5[/tex]∴ The magnitude is 90.5.And,The direction relative to west is[tex]tan\theta = \frac{|A|}{|B|}[/tex]θ = tan⁻¹ (B/A) = tan⁻¹ (64/64) = tan⁻¹ 1 = 45°  south of west∴ The direction is 45° south of west.Again,The magnitude of [tex](\vec{A} - \vec{B})[/tex]: [tex]= \sqrt{|A|^{2}+|B|^{2}}[/tex] [tex]= \sqrt{64^{2}+64^{2}}[/tex] [tex]= 64\sqrt{2}[/tex] [tex]= 90.5[/tex]∴ The magnitude is 90.5.And,The direction relative to west is[tex]tan\theta = \frac{|A|}{|B|}[/tex]θ = tan⁻¹ (B/A) = tan⁻¹ (64/64) = tan⁻¹ 1 = 45° north of west.∴ The direction is 45° north of west.