First slide

Planes in 3D

Question

$The projection of the line \frac{x + 1}{- 1} = \frac{y}{2} = \frac{z - 1}{3} on the plane$ $x - 2 y + z = 6 is the line of intersection of this plane with the plane$

Moderate

A

$2 x + y + 2 = 0$
B

$3 x + y - z = 2$
C

$2 x - 3 y + 8 z = 3$
D

$none of these$

Solution

Equation of the plane through (-1, 0, 1) is
$a (x + 1) + b (y - 0) + c (z - 1) = 0 - - - - - - - (i)$
which is parallel to the given line and perpendicular to the given plane
$\begin{array}{l} - a + 2 b + 3 c = 0 - - - - (ii) \\ and a - 2 b + c = 0 - - - - (iii) \end{array}$
From Eqs. (ii) and (iii), we get
$c = 0, a = 2 b$
$From Eq. (i) 2 b (x + 1) + b y = 0$
$\Rightarrow 2 x + y + 2 = 0$

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