A line with positive direction cosines passes | vedclass

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Question

(B) $1$. Direction Cosines: Since the line makes equal angles with the coordinate axes,its direction ratios are proportional to $(1, 1, 1)$.$2$. Equat

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$\frac{\sqrt{3}}{2}$

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(B) $1$. Direction Cosines: Since the line makes equal angles with the coordinate axes,its direction ratios are proportional to $(1, 1, 1)$.$2$. Equation of the Line: The parametric form of the line passing through $P(2, 1, 2)$ with direction ratios $(1, 1, 1)$ is:$\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-2}{1} = t$Thus,$x = 2+t, y = 1+t, z = 2+t$.$3$. Point of Intersection with Plane: Substituting these into the plane equation $2x+y+z=9$:$2(2+t) + (1+t) + (2+t) = 9$$4 + 2t + 1 + t + 2 + t = 9$$4t + 7 = 9 \Rightarrow 4t = 2 \Rightarrow t = \frac{1}{2}$.$4$. Coordinates of $Q$: Substituting $t = \frac{1}{2}$ back,we get $Q = (2+\frac{1}{2}, 1+\frac{1}{2}, 2+\frac{1}{2}) = (\frac{5}{2}, \frac{3}{2}, \frac{5}{2})$.$5$. Length of $PQ$: Using the distance formula:$PQ = \sqrt{(\frac{5}{2}-2)^2 + (\frac{3}{2}-1)^2 + (\frac{5}{2}-2)^2}$$PQ = \sqrt{(\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ 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 $PQ$ equals $\qquad$ units.

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