The correct option is A 6 We can write =∑((a j)i (i j)a)^2 =∑((a j)i)^2 =((a j)i)^2+((a k)j)^2+((a i)k)^2 =((a_2)i)^2+((a_3)j)^2+((a_21 ...

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Properties of Conjugate of a Complex Number

If a⃗=î+ĵ-2...

Question

If $\to a =^i +^j - 2^k$ , then $∑{(→a×^i)×^j}2=$

6

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36

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\sqrt{6}

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6 \sqrt{6}

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Solution

The correct option is A $6$
We can write
$= \sum ((¯ ¯ ¯ a - ¯ j) ¯ i - (¯ i - ¯ j) ¯ ¯ ¯ a)^{2}$
$= \sum ((¯ ¯ ¯ a - ¯ j) ¯ i)^{2}$
$= ((¯ ¯ ¯ a - ¯ j) ¯ i)^{2} + ((¯ ¯ ¯ a - ¯ ¯ ¯ k) ¯ j)^{2} + ((¯ ¯ ¯ a - ¯ i) ¯ ¯ ¯ k)^{2}$
$= ((a_{2}) ¯ i)^{2} + ((a_{3}) ¯ j)^{2} + ((a_{2} 1 ¯ ¯ ¯ k)^{2}$
$= a_{2}^{2} + a_{3}^{2} + a_{1}^{2}$
$= | ¯ ¯ ¯ a |^{2}$
$= 6$

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Similar questions

\to a =^i +^j

\to b = -^j + 2^k

(\to a - \to b) . (\to a + 2 \to b)

\to a = 2^i +^j +^k

\to b =^i + 2^j + 2^k

\to c =^i +^j + 2^k

\to a \times (\to b \times \to c) = (1 + α)^i + β (1 + α)^j + r (1 + α) (1 + β)^k

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(\to a - 2 \to b) . (\to a + \to b)

\to a = 3^i + 4^j - 5^k

|^i \times (\to a \times^i) +^j \times (\to a \times^j) +^k \times (\to a \times^k) | =

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\to b = 2^i - 4^j - 3^k, \to c = -^i + 2^j + 2^k

\to a + \to b + \to c = 4^i - 4^j

\to a =^i -^j + 2^k, \to b = 2^i + 3^j +^k, \to c =^i -^k

\to a + 2 \to b - 3 \to c

\sqrt{78}

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