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Question: How do you find the magnitude and direction angle of the vector \[v = 6i - 6j\]?...

How do you find the magnitude and direction angle of the vector v=6i6jv = 6i - 6j?

Explanation

Solution

Hint : First we will consider θ\theta as the direction angle of the vector. Then compare the given vector with the general form of vector and then accordingly evaluate the values. Evaluate the magnitude of the vector and then the direction angle of the vector.

Complete step-by-step answer :
We will start off by mentioning the general formula of vectors v=ai+bj\mathop v\limits^ \to = a\mathop i\limits^ \to + b\mathop j\limits^ \to . Now we will compare the given equation with the general form of vectors.
Hence, the values are
a=6 b=6   a = 6 \\\ b = - 6 \;
Now we will evaluate the magnitude of the triangle with the help of v2=a2+b2|\mathop v\limits^ \to {|^2} = {a^2} + {b^2} .

v2=a2+b2 v2=62+(6)2 v=62 {|\mathop v\limits^ \to {|^2}} = {{a^2} + {b^2}} \\\ {|\mathop v\limits^ \to {|^2}} = {{6^2} + {{\left( { - 6} \right)}^2}} \\\ {\mathop v\limits^ \to } = {6\sqrt 2 }

Hence, the magnitude of the given vector is 626\sqrt 2 .
As we know that the sides with lengths a,ba,b\, and v|\mathop v\limits^ \to | form a right-angle triangle.
Now we consider an angle in the right-angled triangle as θ\theta .
Hence, we can write,
sinθ=av cosθ=bv   \sin \theta = \dfrac{a}{{|\mathop v\limits^ \to |}} \\\ \cos \theta = \dfrac{b}{{|\mathop v\limits^ \to |}} \;
Now we will substitute values in the above terms.
sinθ=bv sinθ=662 sinθ=12 cosθ=av cosθ=662 cosθ=12   {\sin \theta } = {\dfrac{b}{{|\mathop v\limits^ \to |}}} \\\ {\sin \theta } = {\dfrac{{ - 6}}{{6\sqrt 2 }}} \\\ {\sin \theta } = {\dfrac{{ - 1}}{{\sqrt 2 }}} \\\ {\cos \theta } = {\dfrac{a}{{|\mathop v\limits^ \to |}}} \\\ {\cos \theta } = {\dfrac{6}{{6\sqrt 2 }}} \\\ {\cos \theta } = {\dfrac{1}{{\sqrt 2 }}} \;
Now, as we know that sine function is negative in the fourth quadrant while the cosine function is positive. So here the angle must be 450- {45^0} .
Hence, the vector has magnitude 626\sqrt 2 and the direction angle of the vector is 450- {45^0} .
So, the correct answer is “The vector has magnitude 626\sqrt 2 and the direction angle of the vector is 450- {45^0} ”.

Note : While comparing the values of the terms with the general equation make sure you compare the values along with their respective signs. Remember that the formula for the magnitude of any vector is given by v2=a2+b2|\mathop v\limits^ \to {|^2} = {a^2} + {b^2} . Also, remember that the sine function is positive in only the first quadrant and second quadrant and the cosine function is positive in the first quadrant and fourth quadrant.