If r¯ satisfies the equation r×i¯+2j¯+k¯= i¯-k¯, then for any scalar m, r¯ is equal to

Let r=xi+yj+zkr×(y+2j+k)=ijkxyz121 =i(y-2z)-j(x-z)+k(2x-y) r×(i+2j+k)=i-k y-2z=1 x-z=0 2x-y=1 x=z y=2x+1 r=xi+(2x+1)j+xk =j+x(i+2j+k)

If r¯ satisfies the equation r×i¯+2j¯+k¯= i¯-k¯, then for any scalar m, r¯ is equal to

Let r=xi+yj+zkr×(y+2j+k)=ijkxyz121 =i(y-2z)-j(x-z)+k(2x-y) r×(i+2j+k)=i-k y-2z=1 x-z=0 2x-y=1 x=z y=2x+1 r=xi+(2x+1)j+xk =j+x(i+2j+k)

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$If \bar{r} satisfies the equation r \times (\bar{i} + 2 \bar{j} + \bar{k}) = \bar{i} - \bar{k}, then for any scalar m, \bar{r} is equal to$

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$\bar{i} + m (\bar{i} + 2 \bar{j} + \bar{k})$

$\bar{k} + m (\bar{i} + 2 \bar{j} + \bar{k})$

$\bar{j} + m (\bar{i} + 2 \bar{j} + \bar{k})$

$\bar{i} - \bar{k} + m (\bar{i} + 2 \bar{j} + \bar{k})$

answer is C.

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Detailed Solution

Let $r = xi + yj + zk$

$r \times (y + 2 j + k) = |\begin{matrix} i & j & k \\ x & y & z \\ 1 & 2 & 1 \end{matrix}| = i (y - 2 z) - j (x - z) + k (2 x - y) r \times (i + 2 j + k) = i - k y - 2 z = 1 x - z = 0 2 x - y = 1 x = z y = 2 x + 1 r = xi + (2 x + 1) j + xk = j + x (i + 2 j + k)$