The line passing through (1, 1, -1) and paral | vedclass

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(B) The line passing through $(1, 1, -1)$ and parallel to the vector $\hat{i} + 2 \hat{j} - \hat{k}$ is given by: $\frac{x - 1}{1} = \frac{y - 1}{2} =

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$2 \sqrt{6}$

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(B) The line passing through $(1, 1, -1)$ and parallel to the vector $\hat{i} + 2 \hat{j} - \hat{k}$ is given by: $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z + 1}{-1} = k$ So,$x = k + 1, y = 2k + 1, z = -k - 1$. To find the point $A$ where this line meets $\frac{x - 3}{-1} = \frac{y + 2}{5} = \frac{z - 2}{-4}$,substitute the parametric coordinates into the second line equation: $\frac{k + 1 - 3}{-1} = \frac{2k + 1 + 2}{5} \Rightarrow \frac{k - 2}{-1} = \frac{2k + 3}{5} \Rightarrow 5k - 10 = -2k - 3 \Rightarrow 7k = 7 \Rightarrow k = 1$. Thus,$A = (1 + 1, 2(1) + 1, -1 - 1) = (2, 3, -2)$. To find the point $B$ where the line meets the plane $2x - y + 2z + 7 = 0$,substitute the parametric coordinates into the plane equation: $2(k + 1) - (2k + 1) + 2(-k - 1) + 7 = 0$ $2k + 2 - 2k - 1 - 2k - 2 + 7 = 0 \Rightarrow -2k + 6 = 0 \Rightarrow k = 3$. Thus,$B = (3 + 1, 2(3) + 1, -3 - 1) = (4, 7, -4)$. The distance $AB$ is: $|AB| = \sqrt{(4 - 2)^2 + (7 - 3)^2 + (-4 - (-2))^2} = \sqrt{2^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$.

The line passing through $(1, 1, -1)$ and parallel to the vector $\hat{i} + 2 \hat{j} - \hat{k}$ meets the line $\frac{x - 3}{-1} = \frac{y + 2}{5} = \frac{z - 2}{-4}$ at $A$ and the plane $2 x - y + 2 z + 7 = 0$ at $B$. Then $AB = $

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