Question Video: Finding a Vector Given Its Norm and Direction Cosines | Nagwa Question Video: Finding a Vector Given Its Norm and Direction Cosines | Nagwa

Question Video: Finding a Vector Given Its Norm and Direction Cosines Mathematics • Third Year of Secondary School

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Find the vector 𝐀 of norm 61 and direction cosines (1/2, βˆ’1/2, √(2)/2).

02:23

Video Transcript

Find the vector 𝐀 of norm 61 and direction cosines one-half, negative one-half, and root two over two.

Let’s begin by reminding ourselves of the definition of direction cosines for a vector 𝐀. For a vector 𝐀 with components 𝐴 π‘₯, 𝐴 𝑦, and 𝐴 𝑧, the direction angles are the angles πœƒ π‘₯, πœƒ 𝑦, and πœƒ 𝑧 that the vector 𝐀 makes with the π‘₯-, 𝑦-, and 𝑧-axes, respectively. The direction cosines are the cosines of the direction angles, where we recall that the norm of the vector is its magnitude.

Now, recalling also that in a right angle triangle, the cosine of angle πœƒ is the length of the side adjacent to the angle divided by the length of the hypotenuse. In our case then, the cos of direction angle πœƒ π‘₯ is given by the π‘₯-component of vector 𝐀 divided by the magnitude of vector 𝐀 and similarly for the cosines of direction angles πœƒ 𝑦 and πœƒ 𝑧. And we write our direction cosines in component form as shown.

Now, we’ve been given the direction cosines and the norm or magnitude of our vector 𝐀. And we’re going to use these to find the components 𝐴 π‘₯, 𝐴 𝑦, and 𝐴 𝑧 of our vector 𝐀. Now, looking at our three direction cosine equations, taking cos πœƒ π‘₯, for instance, multiplying through by the norm of 𝐀, we find the norm of 𝐀 multiplied by cos πœƒ π‘₯ is equal to the π‘₯-component of our vector 𝐀. And applying the same process for our 𝑦- and 𝑧-components, we have our three components of vector 𝐀 in terms of the direction cosines and the norm.

We’re given the direction cosines cos πœƒ π‘₯ is one-half, cos πœƒ 𝑦 is negative a half, and cos πœƒ 𝑧 is root two over two and that the norm is equal to 61. And substituting these values into our equations for the components 𝐴 π‘₯, 𝐴 𝑦, and 𝐴 𝑧, we have 61 multiplied by a half is equal to 𝐴 π‘₯, 61 multiplied by negative a half is 𝐴 𝑦, and 61 multiplied by root two over two is equal to 𝐴 𝑧.

Hence, the vector with norm 61 and direction cosines one-half, negative a half, and root two over two has components 61 over two, negative 61 over two, and 61 root two over two.

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