# Question Video: Finding the Algebraic Form of a Vector That Makes Equal Angles with the Axes Given Its Norm Mathematics • 12th Grade

Find the algebraic form of a vector π if its norm is 31, given that it makes equal angles with the positive directions of the Cartesian axes.

02:52

### Video Transcript

Find the algebraic form of a vector π if its norm is 31, given that it makes equal angles with the positive directions of the Cartesian axes.

Letβs begin by recalling that the algebraic, also called the Cartesian, form of a vector π with components π΄ π₯, π΄ π¦, and π΄ π§ is given by the π₯-component π΄ π₯ multiplied by the unit vector π’ in the positive π₯-direction plus the π¦-component π΄ π¦ multiplied by the unit vector π£ in the positive π¦-direction plus the π§-component π΄ π§ multiplied by the unit vector π€ in the positive π§-direction. And so, to find the algebraic from of our vector π, we need to find the components π΄ π₯, π΄ π¦, and π΄ π§.

The information weβve been given is that the norm of the vector is 31. That is, its magnitude is 31. Weβre also told that π makes equal angles with the positive directions of the Cartesian axes. And this tells us that our direction angles π π₯, π π¦, and π π§ are all equal. And remember that these are the angles the vector makes with the positive π₯-, π¦-, and π§-axes. If π π₯, π π¦, and π π§ are all the same, letβs just call this angle π. Now, recall also that the direction cosines are the cosines of the direction angles, that is, the cos of π π₯, the cos of π π¦, and the cos of π π§.

And using right angle trigonometry, we can show that the components of the vector π, π΄ π₯, π΄ π¦, and π΄ π§, are given by the magnitude or norm of the vector π multiplied by the direction cosines. And in our case, since our direction angles are all equal, that is, π, then our three components π΄ π₯, π΄ π¦, and π΄ π§ are equal to the magnitude or norm of the vector multiplied by cos π. And since our norm is 31, thatβs 31 cos π.

Now, in order to find π, we can use the known result that the sum of the squares of our direction cosines is equal to one. And in our case, since our three angles are equal, this tells us that three cos squared π is equal to one. And now dividing both sides by three and taking the positive and negative square roots, we have the cos of our angle π is positive or negative one over root three. And rationalizing our denominator, this gives us positive or negative root three over three.

In our equation for the components of our vector then, we have π΄ π₯ is equal to π΄ π¦ is equal to π΄ π§ is equal to positive or negative 31 root three over three. And so, in algebraic or Cartesian form, the vector π with norm 31 and making equal angles with the positive directions of the Cartesian axes is positive or negative 31 times the square root of three over three multiplied by the sum of the unit vectors π’, π£, and π€.