Find the equation of the plane that passes through (1,0,0) and is perpendicular to the line r=<1,0,2>+t<1,0,1>. The vector equation of a plane is n*(r-rsub0)=0. Thank you so much for your help!
To find the equation of the plane that passes through a given point and is perpendicular to a given line, we need to follow these steps:
Step 1: Find the normal vector of the plane.
Step 2: Plug in the values of the given point and the normal vector into the vector equation of the plane.
Now let's go step by step:
Step 1: Find the normal vector of the plane.
First, we need to find a vector that is perpendicular to the line given by r = <1,0,2> + t<1,0,1>. This can be done by finding the cross product of any two non-parallel vectors on the line.
Let's take two points on the line, say A(1, 0, 2) and B(2, 0, 3), and subtract their coordinates to get a vector in the direction of the line, which is <1, 0, 1>.
Cross product: <1, 0, 1> x <1, 0, 1>
To calculate the cross product, we can use the formula:
<1, 0, 1> x <1, 0, 1> = <(0*1 - 1*0), (1*1 - 1*0), (1*0 - 0*1)>
= <0, 1, 0>
Therefore, the normal vector of the plane is <0, 1, 0>.
Step 2: Plug in the values of the given point and the normal vector into the vector equation of the plane.
Using the equation n * (r - r0) = 0, where n is the normal vector, r is the position vector, and r0 is the given point, we can now substitute in the values.
The equation of the plane is:
<0, 1, 0> * (r - <1, 0, 0>) = 0
Distributing the dot product:
0*(x - 1) + 1*(y - 0) + 0*(z - 0) = 0
Simplifying further:
(y - 0) = 0
The equation of the plane that passes through (1,0,0) and is perpendicular to the line r = <1,0,2> + t<1,0,1> simplifies to:
y = 0.
So, the equation of the plane is y = 0.
That's it! The equation of the plane that passes through (1,0,0) and is perpendicular to the line r = <1,0,2> + t<1,0,1> is y = 0.