Question

find the direction cosines of the vector i^ +2j^+3k^

Answer 1

$\textbf{Given vector is \overrightarrow{i}+2\,\overrightarrow{j}+3\,\overrightarrow{k} }$

$\textbf{To find:}$

$\text{Direction cosines of \overrightarrow{i}+2\,\overrightarrow{j}+3\,\overrightarrow{k} }$

$\textbf{Solution:}$

$\text{We know that,}$

$\text{The direction cosines of the vector \overrightarrow{r}=x\,\overrightarrow{i}+y\,\overrightarrow{j}+z\,\overrightarrow{k} is}$

$\bf(\dfrac{x}{\sqrt{x^2+y^2+z^2}},\dfrac{y}{\sqrt{x^2+y^2+z^2}},\dfrac{z}{\sqrt{x^2+y^2+z^2}})$

$\text{Then, the direction cosines of \overrightarrow{i}+2\,\overrightarrow{j}+3\,\overrightarrow{k} is}$

$(\dfrac{1}{\sqrt{1^2+2^2+3^2}},\dfrac{2}{\sqrt{1^2+2^2+3^2}},\dfrac{3}{\sqrt{1^2+2^2+3^2}})$

$(\dfrac{1}{\sqrt{1+4+9}},\dfrac{2}{\sqrt{1+4+9}},\dfrac{3}{\sqrt{1+4+9}})$

$(\dfrac{1}{\sqrt{14}},\dfrac{2}{\sqrt{14}},\dfrac{3}{\sqrt{14}})$

$\therefore\textbf{The direction cosines of \bf\overrightarrow{i}+2\,\overrightarrow{j}+3\,\overrightarrow{k} is \bf(\dfrac{1}{\sqrt{14}},\dfrac{2}{\sqrt{14}},\dfrac{3}{\sqrt{14}}) }$