A-a-a-a-a-k-a-k- is always equal to

A⃗=a_1î+a_2ĵ+a_3k̂ ⇒a⃗.î=a_1, a⃗.ĵ=a_2, a⃗.k̂=a_3 a⃗×î=(a_1 î+a_2 ĵ+a_3 k̂)×î= a_2 k̂+ a_3 ĵ Similarly, a⃗×ĵ=a_1 k̂ a_3 î, a⃗×k̂= a_1 ĵ+a_2 î ⇒ (a⃗. ...

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Byju's Answer

Standard XI

Mathematics

Addition of Vectors

a⃗.îa⃗×î+a⃗.ĵ...

Question

$(\to a .^i) (\to a \times^i) + (\to a .^j) (\to a \times^j) + (\to a .^k) (\to a \times^k)$ is always equal to

$\to a$

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$2 \to a$

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$3 \to a$

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None of these

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Solution

The correct option is D
None of these

$\to a = a_{1}^i + a_{2}^j + a_{3}^k \Rightarrow \to a .^i = a_{1}, \to a .^j = a_{2}, \to a .^k = a_{3}$
$\to a \times^i = (a_{1}^i + a_{2}^j + a_{3}^k) \times^i = - a_{2}^k + a_{3}^j$
Similarly, $\to a \times^j = a_{1}^k - a_{3}^i, \to a \times^k = - a_{1}^j + a_{2}^i \Rightarrow (\to a .^i) (\to a \times^i) + (\to a .^j) (\to a \times^j) + (\to a .^k) (\to a \times^k) = - a_{1} a_{2}^k + a_{1} a_{3}^j + a_{1} a_{2}^k - a_{3} a_{2}^i - a_{3} a_{1}^j + a_{3} a_{2}^i = \to 0$

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(→a.^i)(→a×^i)+(→a.^j)(→a×^j)+(→a.^k)(→a×^k) is always equal to

$(\to a .^i) (\to a \times^i) + (\to a .^j) (\to a \times^j) + (\to a .^k) (\to a \times^k)$ is always equal to