The distance of the point (-1, 2, 3) from the | vedclass

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(C) Let the lines be $L_1: \vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{k})$ and $L_2: \vec{r} = (2\hat{i} - \hat{j}) + \mu(\hat{i} - \hat{

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$2\sqrt{6}$

Vedclass · Answer

(C) Let the lines be $L_1: \vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{k})$ and $L_2: \vec{r} = (2\hat{i} - \hat{j}) + \mu(\hat{i} - \hat{j} + \hat{k})$.The direction vector of the line of shortest distance is given by the cross product of the direction vectors of $L_1$ and $L_2$:$\vec{n} = (2\hat{i} + \hat{k}) 	imes (\hat{i} - \hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & 1 \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(0 - (-1)) - \hat{j}(2 - 1) + \hat{k}(-2 - 0) = \hat{i} - \hat{j} - 2\hat{k}$.The line passing through the point $P(-1, 2, 3)$ and parallel to $\vec{n}$ is given by $\frac{x+1}{1} = \frac{y-2}{-1} = \frac{z-3}{-2} = r$.Any point on this line is $(r-1, 2-r, 3-2r)$.To find the intersection with the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 10$,we substitute the coordinates into the plane equation:$(r-1) - 2(2-r) + 3(3-2r) = 10$$r - 1 - 4 + 2r + 9 - 6r = 10$$-3r + 4 = 10 \Rightarrow -3r = 6 \Rightarrow r = -2$.The intersection point $Q$ is $(-2-1, 2-(-2), 3-2(-2)) = (-3, 4, 7)$.The distance $PQ = \sqrt{(-3 - (-1))^2 + (4 - 2)^2 + (7 - 3)^2} = \sqrt{(-2)^2 + 2^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.

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