Question

Let \[\vec{a} = 2 \vec{i} + 3 \vec{j} + 4 \vec{k} \text { and } \vec{b} = 3 \vec{i} + 4 \vec{j} + 5 \vec{k}\] Find the angle between them.

Answer 1

We have:

\[\vec{a} = 2 \vec{i} + 3 \vec{j} + 4 \vec{k} \]

\[ \vec{b} = 3 \vec{i} + 4 \vec{j} + 5 \vec{k} \]

Using scalar product, we can find the angle between vectors \[\vec{a}\] and \[\vec{b}\].

i.e.,

\[\vec{a} . \vec{b} = \left| \vec{a} \right|\left| \vec{b} \right| \cos \theta\]

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

\[= \cos^{- 1} \left( \frac{2 \times 3 + 3 \times 4 + 4 \times 5}{\sqrt{\left( 2^2 + 3^2 + 4^2 \right)} \sqrt{\left( 3^2 + 4^2 + 5^2 \right)}} \right)\]

\[ = \cos^{- 1} \left( \frac{38}{\sqrt{29} \sqrt{50}} = \cos^{- 1} \frac{38}{\sqrt{1450}} \right)\]

∴ The required angle is \[\cos^{- 1} \frac{38}{\sqrt{1450}} .\]