# Find equation of the plane passing through the line $x+y=2, y-z=3$ and is perpendicular to the plane $2x+3y+4z=5$

I am currently stuck on a practice question in my textbook and would appreciate some help on how to solve it.

What i tried: I tried converting the line into parameter form, then taking the cross product between the direction vector and the normal vector of the perpendicular plane, but it does not seem to be the right method. I thought this work since the the cross product of these two vectors should give a normal vector that is perpendicular to both of the other two vectors.

My work: The line into parameter form should be $$(2,0,3)+t(1,-1,1)$$ the cross product between the direction vector $$t (1,-1,1)$$ and the normal vector of the perpendicular plane $$(2,3,4)$$ gives the new normal vector $$N = (7t,2t,-5t)$$. Since the vector $$N$$ should be perpendicular that means I can solve for $$t$$ since the dot product should be 0. $$t$$ ends up being able to be anything, since all the terms in the dot product end up equaling up to $$0$$. Therefore the equation of the plane should be $$7x+2y-5z=0$$.

The solution my text book gives is Textbook solution However I don't really understand what they are doing in the solution here. The part I don't really understand is the first step part that I do not understand. This looks similar to the the parameter form of a line but I don't really understand what this is nor how they got it. From what I understand the equation of a plane is usually of this form Ax+By+Cz=D.

• Since you already have the solution, can you tell us which parts of the solution do you not understand? Dec 31, 2021 at 20:07
• Can you please show your work? You may have made a mistake. Your approach should absolutely work. Textbook solution relies on the fact that any point on the line should satisfy the equation of the plane as the line is in the plane. Also we know the plane is perpendicular to another given plane. So dot product of their normal vectors would be zero. Dec 31, 2021 at 20:16
• I updated the question to add the part I didn't understand. Dec 31, 2021 at 20:17
• I also added in my work Dec 31, 2021 at 20:35

The line is $$~x = t, y = - t + 2, z = - t - 1$$ and the direction vector of the line is $$\vec d = (1, -1, -1)$$ and not $$(1, -1, 1)$$. That is your first mistake. The desired plane is also perpendicular to plane $$2x + 3y + 4z = 5$$ with normal vector $$\vec n = (2, 3, 4)$$. Cross product of the direction vector of the line contained in the plane and the normal vector of the plane perpendicular to the desired plane would give us the normal vector of the desired plane.

$$\vec n \times \vec d = (1, 6, -5)$$ so the equation of the plane is $$x + 6y - 5z = k$$

As the line $$x + y = 2, y - z = 3$$ is in the plane,

$$x + 6y - 5z = x + y + 5 (y - z) = 2 + 5 \cdot 3 = 17$$

So the equation of the desired plane is $$x + 6y - 5z = 17$$

The book solution simply uses the fact that the equation of the given line should satisfy the equation of the plane as the line is contained in the plane. As the line is $$x + y -2 = 0, y - z - 3 = 0$$, any plane with the equation $$(x + y - 2) + \lambda (y - z - 3) = 0$$ would contain the given line.