Position vector that defines position of vector in three dimensions having formula of

\sqrt{(a^{2} + b^{2} + c^{2})}

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\sqrt{(a + b + c)}

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\sqrt{(a^{2} - b^{2} + c^{2})}

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(a^{2} + b^{2} + c^{2})

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Solution

The correct option is A $\sqrt{(a^{2} + b^{2} + c^{2})}$
Let position vector $\to A = a^i + b^j + c^k$ ,
$| \to A |$ gives the position of position vector.
$\Rightarrow | \to A | = \sqrt{(a^{2} + b^{2} + c^{2})}$

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Similar questions

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

\frac{a^{2} - {(b - c)}^{2}}{{(a + c)}^{2} - b^{2}} + \frac{b^{2} - {(a - c)}^{2}}{{(a + b)}^{2} - c^{2}} + \frac{c^{2} - {(a - b)}^{2}}{{(b + c)}^{2} - a^{2}}

b^{2} c^{2} + c^{2} a^{2} + a^{2} b^{2} > a b c (a + b + c)

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