Answer :
To find the equation of a plane through a given point [tex]\((3, 5, -4)\)[/tex] that is orthogonal to a given line, we need to follow several steps.
### Step 1: Identify the direction vector of the line
The parametric equations of the line are:
[tex]\[ x = -1 + 1t, \quad y = -2 - 3t, \quad z = 5 + 2t \][/tex]
From these equations, we can see that the direction vector of the line is:
[tex]\[ \vec{d} = (1, -3, 2) \][/tex]
### Step 2: Identify the normal vector of the plane
Since the plane is orthogonal to the line, its normal vector [tex]\(\vec{n}\)[/tex] must be the same as the direction vector of the line. Therefore, the normal vector [tex]\(\vec{n}\)[/tex] is:
[tex]\[ \vec{n} = (1, -3, 2) \][/tex]
### Step 3: Write the general form of the plane equation
The general equation of a plane is given by:
[tex]\[ Ax + By + Cz + D = 0 \][/tex]
Where [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] are the components of the normal vector. Given that the normal vector is [tex]\((1, -3, 2)\)[/tex], we have:
[tex]\[ 1x - 3y + 2z + D = 0 \][/tex]
### Step 4: Determine the constant [tex]\(D\)[/tex]
To find the value of [tex]\(D\)[/tex], we substitute the coordinates of the given point [tex]\((3, 5, -4)\)[/tex] into the plane equation:
[tex]\[ 1(3) - 3(5) + 2(-4) + D = 0 \][/tex]
[tex]\[ 3 - 15 - 8 + D = 0 \][/tex]
[tex]\[ -20 + D = 0 \][/tex]
[tex]\[ D = 20 \][/tex]
### Step 5: Write the final equation of the plane
Substituting [tex]\(D = 20\)[/tex] back into the plane equation, we have:
[tex]\[ 1x - 3y + 2z + 20 = 0 \][/tex]
Thus, the equation of the plane is:
[tex]\[ x - 3y + 2z + 20 = 0 \][/tex]
### Step 1: Identify the direction vector of the line
The parametric equations of the line are:
[tex]\[ x = -1 + 1t, \quad y = -2 - 3t, \quad z = 5 + 2t \][/tex]
From these equations, we can see that the direction vector of the line is:
[tex]\[ \vec{d} = (1, -3, 2) \][/tex]
### Step 2: Identify the normal vector of the plane
Since the plane is orthogonal to the line, its normal vector [tex]\(\vec{n}\)[/tex] must be the same as the direction vector of the line. Therefore, the normal vector [tex]\(\vec{n}\)[/tex] is:
[tex]\[ \vec{n} = (1, -3, 2) \][/tex]
### Step 3: Write the general form of the plane equation
The general equation of a plane is given by:
[tex]\[ Ax + By + Cz + D = 0 \][/tex]
Where [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] are the components of the normal vector. Given that the normal vector is [tex]\((1, -3, 2)\)[/tex], we have:
[tex]\[ 1x - 3y + 2z + D = 0 \][/tex]
### Step 4: Determine the constant [tex]\(D\)[/tex]
To find the value of [tex]\(D\)[/tex], we substitute the coordinates of the given point [tex]\((3, 5, -4)\)[/tex] into the plane equation:
[tex]\[ 1(3) - 3(5) + 2(-4) + D = 0 \][/tex]
[tex]\[ 3 - 15 - 8 + D = 0 \][/tex]
[tex]\[ -20 + D = 0 \][/tex]
[tex]\[ D = 20 \][/tex]
### Step 5: Write the final equation of the plane
Substituting [tex]\(D = 20\)[/tex] back into the plane equation, we have:
[tex]\[ 1x - 3y + 2z + 20 = 0 \][/tex]
Thus, the equation of the plane is:
[tex]\[ x - 3y + 2z + 20 = 0 \][/tex]
