Vector Algebra — Class 12 Maths Solution

(i) We have, $\vec a = \hat i + \hat j + \hat k$
$\therefore$ $|\vec a| = \sqrt {{{(1)}^2} + {{(1)}^2} + {{(1)}^2}} = \sqrt {1 + 1 + 1} = \sqrt 3$

(ii) We have, $\vec b = 2\hat i - 7\hat j - 3\hat k$
$\therefore$ $|\vec b| = \sqrt {{2^2} + {{( - 7)}^2} + {{( - 3)}^2}} = \sqrt {4 + 49 + 9} = \sqrt {62}$

(iii) We have, $\vec c = \cfrac{1}{{\sqrt 3 }}\hat i + \cfrac{1}{{\sqrt 3 }}\hat j - \cfrac{1}{{\sqrt 3 }}\hat k$

$\therefore$ $|\vec c| = \sqrt {{{\left( {\cfrac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\cfrac{1}{{\sqrt 3 }}} \right)}^{\vec 2}} + {{\left( { - \cfrac{1}{{\sqrt 3 }}} \right)}^2}}$

$= \sqrt {\cfrac{1}{3} + \cfrac{1}{3} + \cfrac{1}{3}} = \sqrt {\cfrac{3}{3}} = \sqrt 1 = 1$