### Video Transcript

Find the direction angles of vector π is equal to two multiplied by the unit vector π’ minus two times the unit vector π€.

Weβre given a vector π in algebraic or Cartesian form. That is the π₯-component multiplied by π’ plus the π¦-component multiplied by π£ plus the π§-component multiplied by π€, where π’, π£, and π€ are the unit vectors in the positive π₯-, π¦-, and π§-directions. In our case, the π₯-component π΄ π₯ is equal to two. And since the unit vector π£ does not appear in our vector, the π¦-component π΄ π¦ is equal to zero, and the π§-component π΄ π§ is equal to negative two, since we have negative two multiplied by π€, the unit vector in the π§-direction.

Whereas to find the direction angles of the vector π β and we recall that the direction angles are the angles π π₯, π π¦, and π π§ that the vector makes with the π₯-, π¦-, and π§-axes, respectively, and recalling also that the direction cosines are the cosines of these angles, that is, cos π π₯, cos π π¦, and cos π π§ β by right angle trigonometry we know that the cos of π π₯ is equal to the π₯-component divided by the magnitude or norm of the vector π, and similarly, for the cos of π π¦ and for the cos of π π§. And remember also that we know that the magnitude or norm of a vector is the square root of the sum of the squares of the components of the vector.

For a vector π then, substituting the values of the components into the formula for the magnitude, we have the square root of two squared plus zero squared plus negative two squared, which is the square root of eight and which in turn simplifies to two times the square root of two. And making some space and making a note of this, we can then substitute this value with our components into the direction cosines. So, for example, the cos of π π₯ is two over two root two, which is one over root two. And rationalizing the denominator gives us the square root of two over two. Similarly, cos π π¦ is equal to zero over two root two, and thatβs equal to zero. And finally, cos π π§ is negative two over two root two, which is equivalent to negative root two over two.

Now remember, weβre trying to find the direction angles π π₯, π π¦, and π π§. And to do this, we take the inverse cosine of our results. π π₯ is then the inverse cos of root two over two, which is 45 degrees. π π¦ is the inverse cos of zero, and thatβs 90 degrees. Finally, π π§ is the inverse cos of negative root two over two, which is 135 degrees. And we have three direction angles, all between zero and 180 degrees.

Hence, the direction angles of the vector π is equal to two π’ minus two π€, or π π₯ is 45 degrees, π π¦ is 90 degrees, and π π§ is 135 degrees.