Question

find the unit vector parallel to the resultant of the vectors A vector = 2 i^-6j^-3k^ and B vector=4i^+3j^-k^​

Answer 1

Answer:

$\large \text{ \vec A=\left(\dfrac{6}{\sqrt{61}}, \ - \dfrac{3}{\sqrt{61}}, \ - \dfrac{4}{\sqrt{61}}\right) }$

Explanation:

Given :

$\begin{gathered}\large \text{ p=2\hat{i}-6\hat{j}-3\hat k and }\\\\\\\large q=4\hat{i}+3\hat{j}-\hat k\end{gathered}$

First add both vectors

$\begin{gathered}\large \vec p+\vec q=2\hat{i}-6\hat{j}-3\hat k+4\hat{i}+3\hat{j}-\hat k\\\\\\\large \vec p+\vec q=6\hat{i}-3\hat{j}-4\hat k\end{gathered}$

Now finding magnitude

• We know formula for magnitude

$\large \sqrt{x^2+y^2+z^2}$

Put the values here

$\begin{gathered}\large I \ \vec p+\vec q \ I=\sqrt{6^2+(-3)^2+(-4)^2}\\\\\\\large I \ \vec p+\vec q \ I=\sqrt{36+16+9}\\\\\\\large I \ \vec p+\vec q \ I=\sqrt{61}\end{gathered}$

Now for unit vector

$\begin{gathered}\large \text{ \vec A=\dfrac{6\hat{i}-3\hat{j}-4\hat k}{\sqrt{61}} }\\\\\\\large \text{ \vec A=\left(\dfrac{6}{\sqrt{61}}, \ - \dfrac{3}{\sqrt{61}}, \ - \dfrac{4}{\sqrt{61}}\right) }\end{gathered}$