Question

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

Answer 1

We will first write the direction ratios of the given vector which happens to be the coefficients of $\hat i$, $\hat j$ and $\hat k$. Next, find the magnitude of the given vector. Then, write the direction cosines as $\left( {\dfrac{a}{{\left| {\vec v} \right|}},\dfrac{b}{{\left| {\vec v} \right|}},\dfrac{c}{{\left| {\vec v} \right|}}} \right)$, where $a,b$ and $c$ are the direction ratios and $\left| {\vec v} \right|$ is the magnitude of a given vector.

Complete step-by-step answer:
We are the vector is $\hat i + 2\hat j + 3\hat k$
Let $\vec v = \hat i + 2\hat j + 3\hat k$
We will write the direction ratios of the given vector.
Here we have $a = 1,b = 2,c = 3$
We will now find the magnitude of the given vector.
The magnitude of the vector is the square root of the sum of squares of the direction ratios.
Hence, $\left| {\vec v} \right| = \sqrt {{1^2} + {2^2} + {3^2}}$ is the magnitude of the given vector.
On solving the above expression, we get
$\left| {\vec v} \right| = \sqrt {1 + 4 + 9} \\\ \Rightarrow \left| {\vec v} \right| = \sqrt {14} \\\$
Now, the direction cosine of the vector can be calculated as, $\left( {\dfrac{a}{{\left| {\vec v} \right|}},\dfrac{b}{{\left| {\vec v} \right|}},\dfrac{c}{{\left| {\vec v} \right|}}} \right)$, where $a,b$ and $c$ are the direction ratios and $\left| {\vec v} \right|$ is the magnitude of a given vector.
Therefore, for the vector, $\hat i + 2\hat j + 3\hat k$, the direction cosine is $\left( {\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}} \right)$

Note: Direction cosines of a vector are unique. But, direction ratios are not unique, there can be more than one set of direction ratios for a given vector. Also, the sum of squares of sums of the direction ratios is equal to 1.