If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{c}=\hat{j}-\hat{k}, \vec{a} imes \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=1$, then $\vec{b}$

Question

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{c}=\hat{j}-\hat{k}, \vec{a} 	imes \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=1$, then $\vec{b}$

Aicharya · Accepted Answer

Let $\vec{b}=x \hat{i}+y \hat{j}+z \hat{k}$ and we have $\vec{a} \cdot \vec{b}=1$ $\begin{aligned} & 	herefore(\hat{i}+\hat{j}+\hat{k})(x \hat{i}+y \hat{j}+z \hat{k})=1 \Rightarrow x+y+z=1 \ & \vec{a} 	imes \vec{b}=\vec{c} \Rightarrow\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ x & y & z \end{array}ight|=\hat{j}-\hat{k} \ & 	herefore(z-y) \hat{i}-(z-x) \hat{j}+(y-x) \hat{k}=\hat{j}-\hat{k} \ & 	herefore z-y=0, x-z=1, y-x=-1 \ & 	herefore y=z, z=x-1, y=x-1 \ & 	ext { Thus } \vec{b}=x \hat{i}+(x-1) \hat{j}+(x-1) \hat{k} \ & 	herefore x+(x-1)+(x-1)=1 \ & \Rightarrow x=1, y=0, z=0 \ & 	herefore \vec{b}=\hat{i} \end{aligned}$

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{c}=\hat{j}-\hat{k}, \vec{a} \times \vec{b}=\vec{c}$ and $\vec{a}…

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