Answer to Question #131379 in Analytic Geometry for fatima

Question

given that r1=3i-2j+k, r2=2i-4j-3k, r3=-i+2j+2k, the direction of cosines of the vector (r1+r2+r3) is

Answer 1

The Direction Cosines of a vector represent the cosines of the angles that the vector makes with the Co-ordinate Axes. 

They are represented as  , or they can be simply called l, m and n respectively.

To find the Direction Cosines, we first find the Unit-Vector in the Direction of the given vector. Then, each component of that unit-vector simply represents the Direction Cosines.

[To find unit-vector of any given vector, we simply divide the vector by its magnitude]

Here, r1+r2+r3= (3i-2j+k)+(2i-4j-3k)+(-i+2j+2k)= (3+2-1)i + (-2-4+2)j + (1-3+2)k = 4i - 4j + 0k

Let us denote r1+r2+r3 = v

Then , unit vector $\hat{v}=$ $\dfrac{\overrightarrow{v}}{\lvert\overrightarrow{v}\rvert}$ ,

$\;\implies \hat{v} = \dfrac{4i-4j+0k}{\sqrt{4^2+(-4)^2+0^2}} \\ \newline \implies \hat{v} = \dfrac{4i-4j}{\sqrt{16+16}} \\ \newline \implies \hat{v}= \dfrac{4i-4j}{\sqrt{32}} \\ \\ \implies \hat{v} = \dfrac{4i-4j}{4\sqrt{2}} \\ \\ \implies \hat{v} = \dfrac{i-j}{\sqrt{2}} = \left(\dfrac{1}{\sqrt{2}}\right)i + \left(\dfrac{-1}{\sqrt{2}}\right)j+0\;k$

Thus, we have found the unit vector. Now, each component of this unit-vector represents the direction cosines. 

This means that the direction cosines are  $\dfrac{1}{\sqrt{2}}, \dfrac{-1}{\sqrt{2}} and \;0.$