Let A be a point having position vector bar{i | vedclass

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(A) The position vector of any point $P$ on the line is given by $\vec{p} = (1)\bar{i} + (t-3)\bar{j} + (-2t)\bar{k}$.Let the plane be $\pi: \vec{r} \

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$\bar{r} \cdot(-\bar{j}+2 \bar{k})=8$

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(A) The position vector of any point $P$ on the line is given by $\vec{p} = (1)\bar{i} + (t-3)\bar{j} + (-2t)\bar{k}$.Let the plane be $\pi: \vec{r} \cdot (2\bar{i} + 3\bar{j} + 5\bar{k}) = 0$. The distance $d$ of point $P$ from the plane is $d = \frac{|(1)(2) + (t-3)(3) + (-2t)(5)|}{\sqrt{2^2 + 3^2 + 5^2}} = \frac{|2 + 3t - 9 - 10t|}{\sqrt{4+9+25}} = \frac{|-7t - 7|}{\sqrt{38}}$.For minimum distance,we set the numerator to zero: $-7t - 7 = 0 \implies t = -1$.Substituting $t = -1$ into the line equation,we get the position vector of $P$: $\vec{p} = \bar{i} + (-1-3)\bar{j} - 2(-1)\bar{k} = \bar{i} - 4\bar{j} + 2\bar{k}$.The vector $\vec{AP} = \vec{p} - \vec{a} = (\bar{i} - 4\bar{j} + 2\bar{k}) - (\bar{i} - 3\bar{j}) = -\bar{j} + 2\bar{k}$.The equation of the plane passing through $P(\bar{i} - 4\bar{j} + 2\bar{k})$ and perpendicular to $\vec{AP} = -\bar{j} + 2\bar{k}$ is $\vec{r} \cdot (-\bar{j} + 2\bar{k}) = \vec{p} \cdot (-\bar{j} + 2\bar{k})$.Calculating the dot product: $(\bar{i} - 4\bar{j} + 2\bar{k}) \cdot (-\bar{j} + 2\bar{k}) = (-4)(-1) + (2)(2) = 4 + 4 = 8$.Thus,the equation is $\bar{r} \cdot (-\bar{j} + 2\bar{k}) = 8$.

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