First slide

Straight lines in 3D

Question

In R³, let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes $P_{1} : 2 x - y + z - 1 = 0$ . Let M be the locus of the feet of the perpendicular drawn from the points on L to the plane P₁. Which of the following points lie(s) on M?

Moderate

A

$(0, - \frac{5}{6}, \frac{- 2}{3})$
B

$(- \frac{1}{6}, - \frac{1}{3}, \frac{1}{6})$
C

$(- \frac{5}{6}, 0, \frac{1}{6})$
D

$(- \frac{1}{3}, 0, \frac{2}{3})$

Solution

Line L will be parallel to the line of intersection of P₁ and P₂ a,b,c are the dr's of line L.
$a + 2 b - c = 0, 2 a - b + c = 0$
$a : b : c = 1 : - 3 : - 5$
‘L’ is $\frac{x - 0}{1} = \frac{y - 0}{- 3} = \frac{z - 0}{- 5}$
foot of the perpendicular from origin to the plane P₁ is $(- \frac{1}{6}, \frac{- 1}{3}, \frac{1}{6})$
equation of projection of the line L on plane P₁, is $\frac{x + \frac{1}{6}}{1} = \frac{y + \frac{2}{6}}{- 3} = \frac{z - \frac{1}{6}}{- 5} = k$
$(0, - \frac{5}{6}, \frac{- 2}{3}) (\frac{- 1}{6}, \frac{- 1}{3}, \frac{1}{6})$ satisfy M

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