Question

Find the unit vector in the direction of the sum of the vectors, \( \vec{a} =2 \hat{i}+ 2 \hat{j}- 5 \hat{k} \) and \( \vec{b} =2 \hat{i}+ \hat{j} + 3 \hat{k} \)

Answer 1

Given vector is a \( \vec{a} =2 \hat{i}+ 2 \hat{j}- 5 \hat{k} \) and \( \vec{b} =2 \hat{i}+ \hat{j} + 3 \hat{k} \)

Let, \( \vec{c} =( \hat{a}+ \hat{b} ) \)

\( \vec{c} =(2+2) \hat{i}+ (2+1) \hat{j} + (-5+3) \hat{k} \)

\( \vec{c} =4 \hat{i}+3 \hat{j} -2 \hat{k} \)

Magnitude of \( \vec{c} = \sqrt{4^2 +3^2+ (-2)^2} \)

\( |\vec{a}| = \sqrt{16+9+4 } = \sqrt{29} \)

Unit vector in the direction of \( \vec{c} \) is

\( \hat{c}= \dfrac{\vec{c}}{ \text{magnitude of} ~\vec{c}} \)

\( \hat{c}= \dfrac{1}{\sqrt{29}}[4 \hat{i}+3 \hat{j}-2\hat{k}] \)

\( \hat{c}= \dfrac{4 \hat{i} }{\sqrt{29}} + \dfrac{3 \hat{j} }{\sqrt{29}}-\dfrac{2 \hat{k} }{\sqrt{29}} \)

Thus,the required unit vector is
\( \dfrac{4 \hat{i} }{\sqrt{29}} + \dfrac{3 \hat{j} }{\sqrt{29}}-\dfrac{2 \hat{k} }{\sqrt{29}} \)