First slide

Straight lines in 3D

Question

The parametric equation of the line of intersection of the given planes are
$3 x - 6 y - 2 z = 15,$ and $2 x + y - 2 z = 5$ are

Easy

A

$x = t - 1, y = 2 t, z = t + 3$
B

$x = t - \frac{1}{2}, y = 3 t, z = 6 t - 1$
C

$x = 14 t + 3, y = 2 t - 1, z = 15 t$
D

$x = 14 t - 3, y = 2 t + 1, z = 15 t$ $x = 3, y = 1$

Solution

the given planes are $3 x - 6 y - 2 z = 15,$ and $2 x + y - 2 z = 5$
The line of intersection of two planes is parallel to the vector which is cross product of the normal vectors to the planes
hence the vector along the line of intersection of two planes is $|\begin{matrix} i & j & k \\ 3 & - 6 & - 2 \\ 2 & 1 & - 2 \end{matrix}| = i (12 + 2) - j (- 6 + 4) + k (3 + 12)$
This can be simplify as $14 i - 2 j + 15 k$
To get a point on the line of intersection of two planes, substitute $z = 0$ in the plane equations and then solve for the other two variables
it implies that $3 x - 6 y = 15, 2 x + y = 5$
solving the above two simultaneous equations $x = 3, y = 1$
Therefore, the parametric form of the equation of the line of intersection of two planes is $x = 14 t + 3, y = 2 t + 1, z = 15 t$

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