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For given vectors, $ \overrightarrow a=2\hat i − \hat j + 2\hat k$ and $ \overrightarrow b=−\hat i + \hat j − \hat k$, find the unit vector in the direction of the vector $ \overrightarrow a +\overrightarrow b.$

$\begin{array}{1 1}(A) \large \frac{-1}{\sqrt 2}\hat i-\large\frac{1}{\sqrt 2}\hat k \\(B) \large \frac{1}{\sqrt 2}\hat i-\large\frac{1}{\sqrt 2}\hat k \\ (C) \large \frac{-1}{\sqrt 2}\hat i+\large\frac{1}{\sqrt 2}\hat k \\(D) \large \frac{1}{\sqrt 2}\hat i+\large\frac{1}{\sqrt 2}\hat k \end{array} $

1 Answer

  • If $\overrightarrow a=(x_1\hat i+y_1\hat j+z_1\hat k)$ and $\overrightarrow b=(x_2\hat i+y_2\hat j+z_2\hat k)$ then $(\overrightarrow a+\overrightarrow b)=(x_1+x_2)\hat i+(y_1+y_2)\hat j+(z_1+z_2)\hat k$.
  • Unit vector of $\overrightarrow a$ is $\hat a=\large\frac{\overrightarrow a}{|\overrightarrow a|}$
Step 1:
$\overrightarrow a=2\hat i-\hat j+2\hat k$ and $\overrightarrow b=-\hat i+\hat j-\hat k$
Let us find $\overrightarrow a+\overrightarrow b$
$\overrightarrow a+\overrightarrow b=(2-1)\hat i+(-1+1)\hat j+(2-1)\hat k$
$\qquad\quad=\hat i+0\hat j+\hat k$
$\qquad\quad=\hat i+\hat k$
Step 2:
Next let us find the unit vector of $\overrightarrow a+\overrightarrow b$.
The magnitude of $(\overrightarrow a+\overrightarrow b)$ is $|\overrightarrow a+\overrightarrow b|$.
$|\overrightarrow a+\overrightarrow b|=\sqrt{1^2+1^2}$
$\qquad\quad\;\;=\sqrt 2$
Step 3:
Then the unit vector =$\large\frac{(\overrightarrow a+\overrightarrow b)}{|\overrightarrow a+\overrightarrow b|}$
$\qquad\qquad\qquad\;\;\;=\large\frac{\hat i+\hat k}{\sqrt 2}$
$\qquad\qquad\qquad\;\;\;=\large \frac{1}{\sqrt 2}$$\hat i+\large\frac{1}{\sqrt 2}$$\hat k$
answered May 21, 2013 by sreemathi.v

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