The equation of a line passing through (3, -1 | vedclass

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(A) Let the direction ratios of the required line be $a, b, c$.Since the line is perpendicular to the lines with direction vectors $\vec{v}_1 = 2\hat{

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$\frac{x-3}{2} = \frac{y+1}{3} = \frac{z-2}{2}$

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(A) Let the direction ratios of the required line be $a, b, c$.Since the line is perpendicular to the lines with direction vectors $\vec{v}_1 = 2\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{v}_2 = \hat{i} - 2\hat{j} + 2\hat{k}$,we have:$2a - 2b + c = 0$ ... $(1)$$a - 2b + 2c = 0$ ... $(2)$Using the cross product to find the direction ratios $(a, b, c) = \vec{v}_1 	imes \vec{v}_2$:$\frac{a}{(-2)(2) - (1)(-2)} = \frac{b}{(1)(1) - (2)(2)} = \frac{c}{(2)(-2) - (-2)(1)}$$\frac{a}{-4 + 2} = \frac{b}{1 - 4} = \frac{c}{-4 + 2}$$\frac{a}{-2} = \frac{b}{-3} = \frac{c}{-2}$Thus,the direction ratios are proportional to $(2, 3, 2)$.The equation of the line passing through $(3, -1, 2)$ with direction ratios $(2, 3, 2)$ is:$\frac{x - 3}{2} = \frac{y - (-1)}{3} = \frac{z - 2}{2}$$\frac{x - 3}{2} = \frac{y + 1}{3} = \frac{z - 2}{2}$

The equation of a line passing through $(3, -1, 2)$ and perpendicular to the lines $\bar{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$ and $\bar{r} = (2\hat{i} + \hat{j} - 3\hat{k}) + \mu(\hat{i} - 2\hat{j} + 2\hat{k})$ is:

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