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 π of norm 61 and direction cosines (1/2, β1/2, β(2)/2).

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Video Transcript

Find the vector π of norm 61 and direction cosines one-half, negative one-half, and root two over two.

Letβs begin by reminding ourselves of the definition of direction cosines for a vector π. For a vector π with components π΄ π₯, π΄ π¦, and π΄ π§, the direction angles are the angles π π₯, π π¦, and π π§ that the vector π makes with the π₯-, π¦-, and π§-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 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 π₯ is given by the π₯-component of vector π divided by the magnitude of vector π and similarly for the cosines of direction angles π π¦ and π π§. And we write our direction cosines in component form as shown.

Now, weβve been given the direction cosines and the norm or magnitude of our vector π. And weβre going to use these to find the components π΄ π₯, π΄ π¦, and π΄ π§ of our vector π. Now, looking at our three direction cosine equations, taking cos π π₯, for instance, multiplying through by the norm of π, we find the norm of π multiplied by cos π π₯ is equal to the π₯-component of our vector π. And applying the same process for our π¦- and π§-components, we have our three components of vector π in terms of the direction cosines and the norm.

Weβre given the direction cosines cos π π₯ is one-half, cos π π¦ is negative a half, and cos π π§ is root two over two and that the norm is equal to 61. And substituting these values into our equations for the components π΄ π₯, π΄ π¦, and π΄ π§, we have 61 multiplied by a half is equal to π΄ π₯, 61 multiplied by negative a half is π΄ π¦, and 61 multiplied by root two over two is equal to π΄ π§.

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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