pCalculus III: The Dot Product Level 10 of 12 Direction Angles, Direction Cosines, Examples VIIp p The Dot Product Level 10 In this article will go over our second application of the dot product.

Recall that the direction of a nonzero planar vector v can be conveniently described by using the angle theta, measured counterclockwise, from the positive xaxis to the planar vector v. In a similar fashion the direction of a nonzero vector v in space can be conveniently described in terms of the angle between vector v and the three unit vectors i, j, and k, which point along the direction of the positive x, y and z axis respectively. The angle alpha is the angle between vector v and the unit vector i hat. The angle beta is the angle between vector v and the unit vector j hat, and the angle gamma is the angle that vector v makes with the unit vector k hat. These three angles are referred to as the direction angles of vector v. Notice that the direction angles will have values between 0 and pi inclusive, this is the same interval we use for the dot product. To find an expression for the direction angles we simply take vector v and dot it with i hat, j hat, and k hat.

Using the geometric definition of the dot product we obtain the following expressions, and by using the component definition of the dot product we obtain the following expressions. Equating the results from the geometric and component definition we obtain the following, finally we solve for cosine of alpha, beta and gamma, in the first, second and third expression respectively. These expressions are referred to as the direction cosines of vector v. You can easily find the value of the direction angles by using inverse cosine.

Consequently, any nonzero vector v in space has the normalized or unit vector form denoted as follows, by substituting the components of the normalized form with the direction cosines we obtain an alternative version of the normalized from of vector v. This expression says that the direction cosines of vector v are the components of the unit vector in the direction of vector v. This expression essentially represents the scalar projection of vector v along the unit vectors i, j, and k. From this result it follows that cosine of alpha squared plus cosine of beta squared plus cosine of gamma squared is equal to 1 since the normalized form of a vector v represents a unit vector with magnitude equal to 1. With this expression we can see if only two of the direction angles are known, the third direction angle can be found by using this expression. Alright lets go over a couple of examples and illustrate how to find direction angles and direction cosines. Find the direction cosines and direction angles of the vector. Round direction angles to the nearest degree.

Alright here we are given a vector in space written in unit vector form. We are asked to find the direction cosines and direction angles of the vector. Lets first find the magnitude of vector v, using the components of the vector and simplifying we obtain the following for the magnitude of vector v. Next lets use the expressions that we derived earlier and substitute the components of vector v and the magnitude of vector v doing that we obtain the following values for the direction cosines.

From these expressions we can go ahead and find the direction angles by using inverse cosine. Evaluating the expressions and rounding to the nearest degree we obtain the following values for the direction angles of vector v.

These values represent the angle that vector v makes with the standard unit vectors. Also notice that the sum of the square of the direction cosines simplify to 1, this is a quick way to verify that you have the correct direction cosines. Alright lets try the next example. Find the direction cosines and direction angles of the vector.

Round direction angles to the nearest degree. Alright similar to the previous example we are asked to find the direction cosines and direction angles of vector v that has equal positive components. Lets start by finding an expression that represents the magnitude of this vector. Finding the sum of the square of the components and taking the square root we obtain the following expression.

Recall that the square root of a quantity squared is by definition equal to the absolute value of the quantity. But in this problem we are given that the component c is greater than 0 meaning that the value is a positive number so we can rewrite the expression for the magnitude as follows. Next lets go ahead and find the direction cosines. Substituting the components and the magnitude of vector v we obtain the following, and then we simplify obtaining the following expressions for the direction cosines of vector v. Notice that the sum of the square of these values is equal to 1. Lastly lets go ahead and take inverse cosine to solve for the direction angles, evaluating the expression and rounding to the nearest degree we obtain 55 degrees for all three direction angles.

Alright lets go over the final example. If a vector has direction angles alpha and beta, find the third direction angle gamma. Alright here we are given two of the direction angles of a vector in space. We are asked to find gamma, the direction angle that the vector makes with the unit vector k hat.

We can easily find this value by using the relationship cosine of alpha squared plus cosine of beta squared plus cosine of gamma squared is equal to 1. Lets go ahead and substitute these angles into the expression. Next lets go ahead and solve for cosine of gamma squared. Now lets evaluate the expression in the right side, and simplify.

We can now go ahead and take the square root of both sides this will yield to distinct solutions. So we break apart this expression into two separate equations and solve for gamma separately. Taking inverse cosine and evaluating we obtain the following angles for gamma. Alright in our next article we will go over our last application of the dot product and find the work done by constant forces.p

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