# Find the equation of the line, $\vec{L}$ where planes ${P_1}$ and ${P_2}$ intersect:

${P_1}$ is given by $2x-3y+5z=5$
${P_2}$ is given by $-2x-3y+5z=3$

Since the line of intersection lies on both planes, the direction $v$ of the line is orthogonal to the normal vectors of both plane. Hence we can take the direction of the line to be
$v=(2,-3,5)\times (-2,-3,5)=\begin{vmatrix} i & j & k \\ 2 & -3 & 5 \\ -2 & -3 & 5 \end{vmatrix}$
$=i(-15+15)-j(10+10)+k(-6-6)$
$=-20j-12k.$
So $v=-20j-12k$. Next we find a point on the intersection of the two planes. Set $z=0$. Then
$2x-3y=5, \text{and} -2x-3y=3.$
Solving this system of equations we get
$x=\frac{1}{2}, y=-\frac{4}{3}.$
Thus $(\frac{1}{2}, -\frac{4}{3}, 0 )$ lies on both lines. Hence the parametric equation of the intersection line is
$x=\frac{1}{2}, y=-\frac{4}{3}-20t, z= 12 t.$

• Savionf
-1

This type of questions should generally come with a fair bounty. I decided to answer as I thought this might be the first time you are asking a question here :)

• Thank you so much! and yes this is my first time asking a question.

• Let me know if you have any questions about the solution.

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