Let a→=i^+j^+k^, b→=j^-k^. Then a vector r→ satisfying a→×r→=b→ and a→.r→=3 is

a×r=b ⇒ a×(a×r)=a×b 3a-|a|2 r=a×b 3a-3r=ijk11101-1=i(-2)-j(-1)+k(1) 3r=3i+3j+3k-(-2i+j+k)=5i+2j+2k r=53i+23j+23k

Let a→=i^+j^+k^, b→=j^-k^. Then a vector r→ satisfying a→×r→=b→ and a→.r→=3 is

a×r=b ⇒ a×(a×r)=a×b 3a-|a|2 r=a×b 3a-3r=ijk11101-1=i(-2)-j(-1)+k(1) 3r=3i+3j+3k-(-2i+j+k)=5i+2j+2k r=53i+23j+23k

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Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{j} - \hat{k} .$ Then a vector $\vec{r}$ satisfying $\vec{a} \times \vec{r} = \vec{b}$ and $\vec{a} . \vec{r} = 3$ is

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$\frac{5}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}$

$\frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} - \frac{2}{3} \hat{k}$

$\frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}$

$\frac{5}{3} \hat{i} - \frac{2}{3} \hat{j} - \frac{2}{3} \hat{k}$

answer is C.

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Detailed Solution

$a \times r = b \Rightarrow a \times (a \times r) = a \times b 3 a - | a |^{2} r = a \times b 3 a - 3 r = |\begin{matrix} i & j & k \\ 1 & 1 & 1 \\ 0 & 1 & - 1 \end{matrix}| = i (- 2) - j (- 1) + k (1) 3 r = 3 i + 3 j + 3 k - (- 2 i + j + k) = 5 i + 2 j + 2 k r = \frac{5}{3} i + \frac{2}{3} j + \frac{2}{3} k$