Question

Find the direction cosines of the vector \[\hat i + 2\hat j + 3\hat k\].

Answer 1

Hint: Direction cosines: direction cosines of the vector is the \[\cos \] function of the angle made by the vector with the \[x,{\text{ }}y,{\text{ }}z\] axis. So to find the direction cosine of the vector first we have to find the magnitude of that function. Then the ratio of the \[x\] component to the total magnitude of the vector gives the \[\cos \alpha \]. Here \[\alpha \] is the angle made by the vector to the x-axis. Same goes with y and z axes as well.

Complete step-by-step answer:
Given,
A vector \[\hat i + 2\hat j + 3\hat k\]
To find,
The direction cosines of the given vector
Let \[\alpha ,\beta ,\gamma \] be the angle made by \[x,{\text{ }}y,{\text{ }}z\] axis respectively.
So the direction cosines are \[\cos \alpha ,\cos \beta ,\cos \gamma \] of the given vector.
Let \[x = \hat i + 2\hat j + 3\hat k\] is the given vector so,
\[\cos \alpha = \dfrac{{x.i}}{{|x|}}\]
\[\cos \beta = \dfrac{{x.j}}{{|x|}}\] and
\[\cos \gamma = \dfrac{{x.k}}{{|x|}}\]
So, first we find the value of the magnitude of \[x\]
\[|x| = \sqrt {{a^2} + {b^2} + {c^2}} \]
On putting all the values of \[a,{\text{ }}b,{\text{ }}c\]
Here, \[a,{\text{ }}b,{\text{ }}c\] are the \[x,{\text{ }}y,{\text{ }}z\] component of the vector respectively
\[|x| = \sqrt {{1^2} + {2^2} + {3^2}} \]
\[|x| = \sqrt {14} \]
Now we are finding the values of dot product
\[x.i = 1\]
\[x.j = 2\] and
\[x.k = 3\]
On putting all the values in order to determine \[\cos \alpha ,\cos \beta ,\cos \gamma \]
\[\cos \alpha = \dfrac{{x.i}}{{|x|}}\]
On putting the values
\[\cos \alpha = \dfrac{1}{{\sqrt {14} }}\]
\[\cos \beta = \dfrac{{x.j}}{{|x|}}\]
On putting the values
\[\cos \beta = \dfrac{2}{{\sqrt {14} }}\]
\[\cos \gamma = \dfrac{{x.k}}{{|x|}}\]
On putting the values
\[\cos \beta = \dfrac{3}{{\sqrt {14} }}\]
The direction cosines of the given vector \[\hat i + 2\hat j + 3\hat k\] is.
\[\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }}\,and\dfrac{3}{{\sqrt {14} }}\]

Note: The sum of the squares of the direction cosines is 1. \[{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1\]. If you want to check that your answer is right or wrong then go through this equation. In order to find these types of questions first we have to find the value of the magnitude of the vector and after that assume some angels with all the axis and then put the relations of the cosine of that angle that is the ratio of the dot product of the vector with x and the magnitude of the given vector and that are the direction cosines of the vector.