The direction of the vector is 22.62degrees and the modulus of the vector is 13 units
The formula for calculating the magnitude of the vector is expressed as:
[tex]v=\sqrt{v_2^2+v_1^2}[/tex]
Given the coordinate point
v₁ = 12
v₂ = 5
Substitute into the formula above;
[tex]v=\sqrt{5^2+12^2}\\v=\sqrt{25+144}\\v=\sqrt{169}\\v=13[/tex]
Hence the modulus of the vector is 13 units
Get the direction;
[tex]\theta = tan^{-1}\frac{v_2}{v_1}[/tex]
[tex]\theta = tan^{-1}\frac{5}{12} \\\theta = tan^{-1}0.41667\\\theta = 22.62^0[/tex]
Hence the direction of the vector is 22.62degrees and the modulus of the vector is 13 units
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