The point bar{i}-2 bar{j} lies on a line para | vedclass

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(A) The equation of the line passing through $\bar{a} = \bar{i}-2 \bar{j}$ and parallel to $\bar{v} = 2 \bar{i}+\bar{k}$ is $\bar{r} = (\bar{i}-2 \bar

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$-\frac{1}{3}(\bar{i}+6 \bar{j}+2 \bar{k})$

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(A) The equation of the line passing through $\bar{a} = \bar{i}-2 \bar{j}$ and parallel to $\bar{v} = 2 \bar{i}+\bar{k}$ is $\bar{r} = (\bar{i}-2 \bar{j}) + t(2 \bar{i}+\bar{k}) = (1+2t)\bar{i} - 2\bar{j} + t\bar{k}$.The plane passes through $\bar{b} = \bar{i}+2 \bar{j}$ and is parallel to $\bar{u}_1 = 2 \bar{j}-\bar{k}$ and $\bar{u}_2 = \bar{i}+2 \bar{k}$.The normal vector to the plane is $\bar{n} = \bar{u}_1 	imes \bar{u}_2 = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 0 & 2 & -1 \ 1 & 0 & 2 \end{vmatrix} = \bar{i}(4) - \bar{j}(1) + \bar{k}(-2) = 4\bar{i} - \bar{j} - 2\bar{k}$.The equation of the plane is $(\bar{r} - \bar{b}) \cdot \bar{n} = 0$,which is $(\bar{r} - (\bar{i}+2 \bar{j})) \cdot (4\bar{i} - \bar{j} - 2\bar{k}) = 0$.Substituting $\bar{r} = (1+2t)\bar{i} - 2\bar{j} + t\bar{k}$ into the plane equation:$((1+2t-1)\bar{i} + (-2-2)\bar{j} + t\bar{k}) \cdot (4\bar{i} - \bar{j} - 2\bar{k}) = 0$$(2t\bar{i} - 4\bar{j} + t\bar{k}) \cdot (4\bar{i} - \bar{j} - 2\bar{k}) = 0$$8t + 4 - 2t = 0 \implies 6t = -4 \implies t = -\frac{2}{3}$.Substituting $t = -\frac{2}{3}$ into the line equation:$\bar{r} = (1+2(-\frac{2}{3}))\bar{i} - 2\bar{j} + (-\frac{2}{3})\bar{k} = (1-\frac{4}{3})\bar{i} - 2\bar{j} - \frac{2}{3}\bar{k} = -\frac{1}{3}\bar{i} - 2\bar{j} - \frac{2}{3}\bar{k} = -\frac{1}{3}(\bar{i} + 6\bar{j} + 2\bar{k})$.

The point $\bar{i}-2 \bar{j}$ lies on a line parallel to the vector $2 \bar{i}+\bar{k}$. The point $\bar{i}+2 \bar{j}$ lies on a plane parallel to the vectors $2 \bar{j}-\bar{k}$ and $\bar{i}+2 \bar{k}$. Find the point of intersection of the line and the plane.

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