A line with positive direction cosines passes | vedclass

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(C) The line makes equal angles with the coordinate axes,so its direction cosines are equal. Let the direction cosines be $(l, l, l)$. Since $l^2 + l^

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$\sqrt{3}$

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(C) The line makes equal angles with the coordinate axes,so its direction cosines are equal. Let the direction cosines be $(l, l, l)$. Since $l^2 + l^2 + l^2 = 1,$ we have $3l^2 = 1,$ so $l = \frac{1}{\sqrt{3}}$ (as direction cosines are positive).The direction ratios of the line are proportional to $(1, 1, 1).$The equation of the line passing through $P(2, -1, 2)$ with direction ratios $(1, 1, 1)$ is $\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = r.$Any point $Q$ on this line is given by $(r+2, r-1, r+2).$Since $Q$ lies on the plane $2x + y + z = 9,$ we substitute the coordinates of $Q$ into the plane equation:$2(r+2) + (r-1) + (r+2) = 9$$2r + 4 + r - 1 + r + 2 = 9$$4r + 5 = 9$$4r = 4 \Rightarrow r = 1.$Thus,the point $Q$ is $(1+2, 1-1, 1+2) = (3, 0, 3).$The distance $PQ$ is $\sqrt{(3-2)^2 + (0 - (-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}.$

$A$ line with positive direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. If the line meets the plane $2x + y + z = 9$ at point $Q,$ then the length $PQ$ equals

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