Question

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

Answer 1

Hint: First we will consider $ \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 $ \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 \;
  $
Now we will evaluate the magnitude of the triangle with the help of $ |\mathop v\limits^ \to {|^2} = {a^2} + {b^2} $ .
\[
  {|\mathop v\limits^ \to {|^2}} = {{a^2} + {b^2}} \\
  {|\mathop v\limits^ \to {|^2}} = {{6^2} + {{\left( { - 6} \right)}^2}} \\
  {\mathop v\limits^ \to } = {6\sqrt 2 }
\]
Hence, the magnitude of the given vector is $ 6\sqrt 2 $ .
As we know that the sides with lengths $ a,b\, $ and $ |\mathop v\limits^ \to | $ form a right-angle triangle.
Now we consider an angle in the right-angled triangle as $ \theta $ .
Hence, we can write,
 $
  \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.
 $
  {\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 $ - {45^0} $ .
Hence, the vector has magnitude $ 6\sqrt 2 $ and the direction angle of the vector is $ - {45^0} $ .
So, the correct answer is “The vector has magnitude $ 6\sqrt 2 $ and the direction angle of the vector is $ - {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 $ |\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.