The point of intersection of the line passing | vedclass

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(A) The equation of the line passing through $\vec{a} = \hat{i}-\hat{j}$ and $\vec{b} = \hat{j}-\hat{k}$ is given by $\vec{r} = \vec{a} + \lambda(\vec

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$\frac{1}{6}(-5 \hat{i}+16 \hat{j}-11 \hat{k})$

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(A) The equation of the line passing through $\vec{a} = \hat{i}-\hat{j}$ and $\vec{b} = \hat{j}-\hat{k}$ is given by $\vec{r} = \vec{a} + \lambda(\vec{b}-\vec{a})$.$\vec{r} = (\hat{i}-\hat{j}) + \lambda(-\hat{i} + 2\hat{j} - \hat{k}) = (1-\lambda)\hat{i} + (2\lambda-1)\hat{j} - \lambda\hat{k}$ ...$(i)$The equation of the plane passing through $\vec{a} = 2\hat{i}+\hat{j}$,$\vec{b} = 2\hat{j}-\hat{k}$,and $\vec{c} = \hat{i}+2\hat{k}$ is $(\vec{r}-\vec{a}) \cdot [(\vec{b}-\vec{a}) 	imes (\vec{c}-\vec{a})] = 0$.Calculating the normal vector $\vec{n} = (\vec{b}-\vec{a}) 	imes (\vec{c}-\vec{a}) = (-2\hat{i}+\hat{j}-\hat{k}) 	imes (-\hat{i}-\hat{j}+2\hat{k}) = \hat{i} + 5\hat{j} + 3\hat{k}$.So,the plane equation is $(\vec{r} - (2\hat{i}+\hat{j})) \cdot (\hat{i} + 5\hat{j} + 3\hat{k}) = 0$.Substituting $\vec{r}$ from $(i)$ into the plane equation:$((1-\lambda-2)\hat{i} + (2\lambda-1-1)\hat{j} - \lambda\hat{k}) \cdot (\hat{i} + 5\hat{j} + 3\hat{k}) = 0$$(-1-\lambda)\hat{i} + (2\lambda-2)\hat{j} - \lambda\hat{k}) \cdot (\hat{i} + 5\hat{j} + 3\hat{k}) = 0$$(-1-\lambda) + 5(2\lambda-2) - 3\lambda = 0$$-1 - \lambda + 10\lambda - 10 - 3\lambda = 0 \Rightarrow 6\lambda = 11 \Rightarrow \lambda = \frac{11}{6}$.Substituting $\lambda = \frac{11}{6}$ into $(i)$:$\vec{r} = (1-\frac{11}{6})\hat{i} + (2(\frac{11}{6})-1)\hat{j} - \frac{11}{6}\hat{k} = -\frac{5}{6}\hat{i} + \frac{16}{6}\hat{j} - \frac{11}{6}\hat{k} = \frac{1}{6}(-5\hat{i} + 16\hat{j} - 11\hat{k})$.

The point of intersection of the line passing through the points $\hat{i}-\hat{j}$ and $\hat{j}-\hat{k}$ and the plane passing through the points $2 \hat{i}+\hat{j}$,$2 \hat{j}-\hat{k}$,and $\hat{i}+2 \hat{k}$ is

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