First slide

Straight lines in 3D

Question

$If the projection of the line \frac{x}{2} = \frac{y - 1}{2} = \frac{z - 1}{1} on a plane P is \frac{x}{1} = \frac{y - 1}{1} = \frac{z - 1}{- 1} . Then the distance of plane P from origin is$

Moderate

A

$\sqrt{3}$
B

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

$\sqrt{6}$
D

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

Solution

$Projection of line \frac{x}{2} = \frac{y - 1}{2} = \frac{z - 1}{1} on a plane P is$
$\frac{x}{1} = \frac{y - 1}{1} = \frac{z - 1}{- 1}$
Plane through these lines is perpendicular to the plane P Normal to the plane determined by the given lines is
$|\begin{array}{rrr} \hat{i} \hat{j} \hat{k} \\ 2 2 1 \\ 1 1 - 1 \end{array}| = - 3 \hat{i} + 3 \hat{j}$
Direction ratios normal to the required plane is
$Equation of the plane is x + y + 2 z - 3 = 0 as it passes through (0, 1, 1)$
$Its distance from origin is \sqrt{\frac{3}{2}} .$

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