A line with direction cosines passes through | vedclass

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(C) Since the line makes equal angles with the coordinate axes,its direction cosines are $l = m = n$. We know that $l^2 + m^2 + n^2 = 1$,so $3l^2 = 1$

Vedclass · Accepted Answer

$\sqrt{3}$

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(C) Since the line makes equal angles with the coordinate axes,its direction cosines are $l = m = n$. We know that $l^2 + m^2 + n^2 = 1$,so $3l^2 = 1$,which gives $l = m = n = \frac{1}{\sqrt{3}}$. 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 this line is $Q(r+2, r-1, r+2)$. Since $Q$ lies on the plane $2x + y + z = 9$,we substitute the coordinates of $Q$ into the plane equation: $2(r+2) + (r-1) + (r+2) = 9$ $2r + 4 + r - 1 + r + 2 = 9$ $4r + 5 = 9$ $4r = 4 \implies r = 1$. The coordinates of $Q$ are $(1+2, 1-1, 1+2) = (3, 0, 3)$. The length of the line segment $PQ$ is the distance between $P(2, -1, 2)$ and $Q(3, 0, 3)$: $PQ = \sqrt{(3-2)^2 + (0-(-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.

$A$ line with 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 $Q$. Find the length of the line segment $PQ$.

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