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# Find the angle between the vectors : $\overrightarrow a = \hat i - \hat j + \hat k\: and \: \overrightarrow b = \hat i + \hat j -\hat k$

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A)
Toolbox:
• The scalar product of two vectors is $\overrightarrow a.\overrightarrow b=\mid \overrightarrow a\mid\mid\overrightarrow b\mid\cos\theta$
• Hence the angle between two vectors is given by $\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{\mid\overrightarrow a\mid\mid\overrightarrow b\mid}$
Step 1:
$\overrightarrow a=\hat i-\hat j+\hat k$
$\overrightarrow b=\hat i+\hat j-\hat k$
$\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{\mid\overrightarrow a\mid\mid\overrightarrow b\mid}$
$\overrightarrow a.\overrightarrow b=(\hat i-\hat j+\hat k).(\hat i+\hat j-\hat k)$
$\qquad\;=1-1-1$
$\qquad\;=-1$
Step 2:
$\mid\overrightarrow a \mid=\sqrt{1^2+(-1)^2+(1)^2}$
$\mid\overrightarrow a \mid=\sqrt 3$
$\mid\overrightarrow b \mid=\sqrt{1^2+(1)^2+(-1)^2}$
$\mid\overrightarrow b \mid=\sqrt 3$
$\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{\mid\overrightarrow a\mid\mid\overrightarrow b\mid}$
$\qquad=\large\frac{-1}{3}$
$\theta=\cos^{-1}\big(\large\frac{-1}{3}\big)$