Question 200276
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Your vectors are orthogonal to the axes, so the end point of the resultant is simply the *[tex \Large x]-coordinate of your vector *[tex \Large y(7,0)] and the *[tex \Large y]-coordinate of your vector *[tex \Large z(0,6)], in other words  *[tex \Large r(7,6)].  You are correct in saying that the magnitude of the resultant is *[tex \Large \sqrt{85}] because *[tex \Large \sqrt{7^2 + 6^2} = \sqrt{49 + 36} = \sqrt{85}].  The direction of the resultant is the angle the resultant makes with the x-axis or:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{6}{7}\right)]


Read that as <i>"the angle whose tangent is 6 divided by 7"</i>.


Calculator:  First decide whether you want your answer in degrees or radians, then set the calculator mode appropriately.  I choose radians.


Keystrokes:


6


'divide'


7


'='


'inverse function mode'


'tan'


Result: approximately 40.6 radians.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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