A line with positive direction cosines passes | vedclass

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Question

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

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$\sqrt{3}$

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(C) Since the line makes equal angles with the coordinate axes,its direction cosines are equal. Let the direction cosines be $l, m, n$. Since $l=m=n$ and $l^2+m^2+n^2=1$,we have $3l^2=1$,so $l=m=n=\frac{1}{\sqrt{3}}$ (given positive direction cosines).The equation of the line passing through $P(2,-1,2)$ with direction cosines $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ is $\frac{x-2}{1/\sqrt{3}} = \frac{y+1}{1/\sqrt{3}} = \frac{z-2}{1/\sqrt{3}} = t$.Any point on this line is given by $(2+\frac{t}{\sqrt{3}}, -1+\frac{t}{\sqrt{3}}, 2+\frac{t}{\sqrt{3}})$.Since this point lies on the plane $2x+y+z=9$,we substitute the coordinates into the plane equation:$2(2+\frac{t}{\sqrt{3}}) + (-1+\frac{t}{\sqrt{3}}) + (2+\frac{t}{\sqrt{3}}) = 9$$4 + \frac{2t}{\sqrt{3}} - 1 + \frac{t}{\sqrt{3}} + 2 + \frac{t}{\sqrt{3}} = 9$$5 + \frac{4t}{\sqrt{3}} = 9$$\frac{4t}{\sqrt{3}} = 4$$t = \sqrt{3}$.Since $t$ represents the distance $PQ$,the length of the line segment $PQ$ is $\sqrt{3}$.

$A$ 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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