Let a=−i^−j^+k^,a→⋅b→=1 and a→×b→=i^−j^ then a→−6b→ is equal to

Let b→=ai^+bj^+ck^⇒a→⋅b→=1−a−b+c=1a→×b→=ijk−1−11abci-j=i(−c−b)−j(−c−a)+k(−b+a) ⇒−c−b=1,−c−a=1,−b+a=0from −2a+c=1−a-c=1 ⇒3a=-2⇒a=-2/3From (2) -c + 2/3 =1⇒ C=−1/3b→=-2i−2j−k3⇒a→−6b→=-i^−j^+k−2−2i^−2j−k=-i−j+k+4i+4j+2k=3i+3j+3k=3(i^+j^+k^)

Let a=−i^−j^+k^,a→⋅b→=1 and a→×b→=i^−j^ then a→−6b→ is equal to

Let b→=ai^+bj^+ck^⇒a→⋅b→=1−a−b+c=1a→×b→=ijk−1−11abci-j=i(−c−b)−j(−c−a)+k(−b+a) ⇒−c−b=1,−c−a=1,−b+a=0from −2a+c=1−a-c=1 ⇒3a=-2⇒a=-2/3From (2) -c + 2/3 =1⇒ C=−1/3b→=-2i−2j−k3⇒a→−6b→=-i^−j^+k−2−2i^−2j−k=-i−j+k+4i+4j+2k=3i+3j+3k=3(i^+j^+k^)

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Let $a = - \hat{i} - \hat{j} + \hat{k}, \vec{a} \cdot \vec{b} = 1 and \vec{a} \times \vec{b} = \hat{i} - \hat{j} then$ $\vec{a} - 6 \vec{b}$ is equal to

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$3 (\hat{i} + \hat{j} - \hat{k})$

$3 (\hat{i} - \hat{j} + \hat{k})$

$3 (\hat{i} - \hat{j} - \hat{k})$

$3 (\hat{i} + \hat{j} + \hat{k})$

answer is A.

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Detailed Solution

Let $\vec{b} = a \hat{i} + b \hat{j} + c \hat{k} \Rightarrow \vec{a} \cdot \vec{b} = 1$
$\begin{array}{l} - a - b + c = 1 \\ \vec{a} \times \vec{b} = |\begin{array}{ccc} i & j & k \\ - 1 & - 1 & 1 \\ a & b & c \end{array}| \\ i - j = i (- c - b) - j (- c - a) + k (- b + a) \end{array}$

$\Rightarrow - c - b = 1, - c - a = 1, - b + a = 0$
from $- 2 a + c = 1$
$\begin{array}{l} - a - c = 1 \\ \Rightarrow 3 a = - 2 \\ \Rightarrow a = - 2 / 3 \end{array}$
From (2) -c + 2/3 =1
$\begin{array}{l} \Rightarrow C = - 1 / 3 \\ \vec{b} = \frac{- 2 i - 2 j - k}{3} \\ \Rightarrow \vec{a} - 6 \vec{b} = - \hat{i} - \hat{j} + k - 2 (- 2 \hat{i} - 2 j - k) \\ = - i - j + k + 4 i + 4 j + 2 k \\ = 3 i + 3 j + 3 k = 3 (\hat{i} + \hat{j} + \hat{k}) \end{array}$