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Q2  Find the angle between the vectors \hat i - 2 \hat j + 3 \hat k \: \:and \: \: 3 \hat i - 2 \hat j + \hat k

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Given two vectors 

\vec a=\hat i - 2 \hat j + 3 \hat k \: \:and \: \: \vec b=3 \hat i - 2 \hat j + \hat k

Now As we know,

The angle between two vectors \vec a and \vec b is given by 

\theta=cos^{-1}\left ( \frac{\vec a.\vec b}{\left | \vec a \right |\left | \vec b \right |}\right )

Hence the angle between \vec a=\hat i - 2 \hat j + 3 \hat k \: \:and \: \: \vec b=3 \hat i - 2 \hat j + \hat k

\theta=cos^{-1}\left ( \frac{(\hat i-2\hat j+3\hat k).(3\hat i-2\hat j+\hat k)}{\left | \hat i-2\hat j+3\hat k \right |\left |3\hat i-2\hat j+\hat k \right |}\right )

\theta=cos^{-1}\left ( \frac{3+4+3}{\sqrt{1^2+(-2)^2+3^3}\sqrt{3^2+(-2)^2+1^2}} \right )

\theta=cos^{-1}\frac{10}{14}

\theta=cos^{-1}\frac{5}{7}

Posted by

Pankaj Sanodiya

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