Question

overline r =0.2i+aj-0.3k is a unit vector, the value of Ifr a is​

Answer 1

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$if \: \overline r = 0.2i + \: aj - 0.3k \: is \:a \:unit\:vector \: \\ then \\ | \overline r| = 1 \\ \implies \: \sqrt{ {(0.2)}^{2} + {a}^{2} + {(0.3)}^{2} } = 1 \\ \implies \: 0.04 + {a}^{2} + 0.09 = 1 \\ \implies \: {a}^{2} = \sqrt{0.87} \\ \implies \: a = \pm0.93$

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