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

The correct option is C 2 a We have #160; #160;î×( a ×î #160;) =( î .î #160;) a ( a .î #160;) î = a ( a .î #160;) î Simi ...

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Permutation: n Different Things Taken r at a Time

î× a ×î +ĵ× a...

Question

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

\to a

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- \to a

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2 \to a

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- 2 \to a

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(a \cdot^i) (a \times^i) + (a \cdot^j) (a \times^j) + (a \cdot^k) (a \times^k)

\to a = x^i + y^j + z^k

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

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

{∣ ∣^i \times (\to a \times^i) ∣ ∣}^{2} + {∣ ∣^j \times (\to a \times^j) ∣ ∣}^{2} + {∣ ∣^k \times (\to a \times^k) ∣ ∣}^{2}

^i \times ((^a -^j) \times^i) +^j \times ((^a -^k) \times^j) +^k \times ((^a -^i) \times^k) = \to 0

\to a = x^i + y^j + z^k,

8 (x^{3} - x y + z x)

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

{∣ ∣ (\to a \times^i) \times^j ∣ ∣}^{2} + {∣ ∣ (\to a \times^j) \times^k ∣ ∣}^{2} + {∣ ∣ (\to a \times^k) \times^i ∣ ∣}^{2} =

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