Vector Algebra — Class 12 Maths Solution

Let $\vec a = \hat i + 2\hat j + 3\hat k$ and $\vec b = 2\hat i + 3\hat j + \hat k$

Here, $|\vec a| = \sqrt {{{(1)}^2} + {{(2)}^2} + {{(3)}^2}} = \sqrt {1 + 4 + 9} = \sqrt {14}$

and $|\vec b| = \sqrt {{{(2)}^2} + {{(3)}^2} + {{(1)}^2}} = \sqrt {4 + 9 + 1} = \sqrt {14}$

Hence, $\vec a \ne \vec b$ but $|\vec a| = |\vec b|$