$$\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$$ এবং $$|\overrightarrow a | = 1$$

$$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$$ এবং $$|\overrightarrow b | = 1$$

$$\therefore$$ $$a_1^2 + a_2^2 + a_3^2 = 1$$ এবং $$b_1^2 + b_2^2 + b_3^2 = 1$$

আবার $$|\overrightarrow a + \overrightarrow b | = 1$$

বা, $${({a_1} + {b_1})^2} + {({a_2} + {b_2})^2} + {({a_3} + {b_3})^2} = 1$$

বা, $$\left( {a_1^2 + a_2^2 + a_3^2} \right) + \left( {b_1^2 + b_2^2 + b_3^2} \right) + 2\left( {{a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}} \right) = 1$$

বা, $$2({a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}) = - 1$$

$$\therefore$$ $$|\overrightarrow a - \overrightarrow b {|^2} = {({a_1} - {b_1})^2} + {({a_2} - {b_2})^2} + {({a_3} - {b_3})^2}$$

$$ = \left( {a_1^2 + a_2^2 + a_3^2} \right) + \left( {b_1^2 + b_2^2 + b_3^2} \right) - 2({a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3})$$

$$ = 1 + 1 - 2( - 1) = 4$$

$$\therefore$$ $$|\overrightarrow a - \overrightarrow b | = 2$$