If $\vec{a} = \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix}$ and $\vec{c} = \begin{bmatrix} -1 \ -1 \ 0 \end{bmatrix}$, then the vector $\vec{b}$ satisfying $\vec{a} imes \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 1$ is

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If $\vec{a} = \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix}$ and $\vec{c} = \begin{bmatrix} -1 \ -1 \ 0 \end{bmatrix}$, then the vector $\vec{b}$ satisfying $\vec{a} 	imes \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 1$ is

Aicharya · Accepted Answer

Let $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$ But $|\hat{i} - \hat{j} + \hat{k}| \cdot |b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}| = 1$ $\Rightarrow b_1 - b_2 + b_3 = 1$ .......(i) And $\vec{a} 	imes \vec{b} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ b_1 & b_2 & b_3 \end{array}| = -\hat{i}|b_2 + b_3| + \hat{j}|b_1 - b_3| + \hat{k}|b_2 + b_1|$ $\Rightarrow \vec{a} 	imes \vec{b} = \vec{c}$ Comparing the coefficients of $\hat{i}$, $\hat{j}$ and $\hat{k}$ respectively, We get $b_2 + b_3 = 1$ ......(ii) $b_1 - b_3 = -1$ .......(iii) $b_2 + b_1 = 0$ .......(iv) By solving the equations (i), (ii), (iii) and (iv), we get $b_1 = 0$, $b_2 = 0$ and $b_3 = 1$.

If $\vec{a} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$ and $\vec{c} = \begin{bmatrix} -1 \\ -1 \\ 0…

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