A line with positive direction cosines passes | vedclass

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(C) Let the direction cosines of the line be $(l, l, l)$ since it makes equal angles with the coordinate axes. Since $l^2 + l^2 + l^2 = 1$, we have $3

Vedclass · Accepted Answer

$\sqrt{3} 	ext{ unit}$

Vedclass · Answer

(C) Let the direction cosines of the line be $(l, l, l)$ since it makes equal angles with the coordinate axes. Since $l^2 + l^2 + l^2 = 1$, we have $3l^2 = 1$, which gives $l = \frac{1}{\sqrt{3}}$ (taking the positive value). 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 on the line is given by $(r+2, r-1, r+2)$. Since this point lies on the plane $2x+y+z=9$, we substitute the coordinates: $2(r+2) + (r-1) + (r+2) = 9$ $2r + 4 + r - 1 + r + 2 = 9$ $4r + 5 = 9 \Rightarrow 4r = 4 \Rightarrow r = 1$. The point $Q$ is $(1+2, 1-1, 1+2) = (3, 0, 3)$. The length $PQ = \sqrt{(3-2)^2 + (0 - (-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} 	ext{ units}$.

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

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