If $u = i \times (a \times i) + j \times (a \times j) + k \times (a \times k),$ then

  • A
    $u = 0$
  • B
    $u = i + j + k$
  • C
    $u = 2a$
  • D
    $u = a$

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$(D)$ $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{z}-\vec{y})$

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