A line with positive direction cosines passes | vedclass

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(C) Since the line makes equal angles with the coordinate axes,its direction cosines $(l, m, n)$ are equal. Let $l = m = n = a$.Since $l^2 + m^2 + n^2

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$\sqrt{3}$

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(C) Since the line makes equal angles with the coordinate axes,its direction cosines $(l, m, n)$ are equal. Let $l = m = n = a$.Since $l^2 + m^2 + n^2 = 1$,we have $3a^2 = 1$,so $a = \frac{1}{\sqrt{3}}$ (taking positive value as direction cosines are positive).Thus,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:$\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = k$So,any point on the line is given by $(k+2, k-1, k+2)$.Since this point $Q$ lies on the plane $2x+y+z=9$,we substitute the coordinates into the plane equation:$2(k+2) + (k-1) + (k+2) = 9$$2k + 4 + k - 1 + k + 2 = 9$$4k + 5 = 9$$4k = 4 \Rightarrow k = 1$Substituting $k=1$ back into the coordinates of $Q$,we get $Q(1+2, 1-1, 1+2) = Q(3, 0, 3)$.The length of the line segment $PQ$ is the distance between $P(2, -1, 2)$ and $Q(3, 0, 3)$:$PQ = \sqrt{(3-2)^2 + (0 - (-1))^2 + (3-2)^2}$$PQ = \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. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ equals

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