Let vec{a}=hat{i}-hat{j}+2 hat{k} and vec{b} | vedclass

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(A) Given $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$,$\vec{a} 	imes \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$.We know that $|\vec{a} 	imes \

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$\frac{2}{\sqrt{21}}$

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(A) Given $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$,$\vec{a} 	imes \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$.We know that $|\vec{a} 	imes \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} |\vec{b}|^{2}$.Here,$|\vec{a}|^{2} = 1^{2}+(-1)^{2}+2^{2} = 6$.$|\vec{a} 	imes \vec{b}|^{2} = 2^{2}+0^{2}+(-1)^{2} = 5$.Substituting these values,$5+3^{2} = 6 |\vec{b}|^{2} \implies 14 = 6 |\vec{b}|^{2} \implies |\vec{b}|^{2} = \frac{7}{3}$.Now,$|\vec{a}-\vec{b}|^{2} = |\vec{a}|^{2}+|\vec{b}|^{2}-2(\vec{a} \cdot \vec{b}) = 6+\frac{7}{3}-2(3) = \frac{7}{3}$.Thus,$|\vec{a}-\vec{b}| = \sqrt{\frac{7}{3}}$.The projection of $\vec{b}$ on $\vec{a}-\vec{b}$ is given by $\frac{\vec{b} \cdot (\vec{a}-\vec{b})}{|\vec{a}-\vec{b}|}$.$= \frac{\vec{b} \cdot \vec{a} - |\vec{b}|^{2}}{|\vec{a}-\vec{b}|} = \frac{3 - \frac{7}{3}}{\sqrt{\frac{7}{3}}} = \frac{\frac{2}{3}}{\sqrt{\frac{7}{3}}} = \frac{2}{3} 	imes \sqrt{\frac{3}{7}} = \frac{2}{\sqrt{21}}$.

Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :-

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