Find the vector and cartesian equations of a plane passing through the points (2,2,-1), (3,4,2) and (7,0,6). Also, find the vector equation of a plane passing through (4,3,1) and parallel to the plane mentioned above.

Or

Find the vector equation of the plane that contains the line and the point (-1,3,-4). Also, find the length of the perpendicular drawn from the point (2,1,4) to the plane thus obtained.

Let A≡(2,2,-1), B≡(3,4,2) and C≡(7,0,6).

If the co-ordinates of points A≡ (x₁, y₁, z₁) and B≡ (x₂, y₂, z₂), then the vector is

Hence, and

We have to figure out the normal vector of the required plane. Since, and lie on the plane, therefore will be perpendicular to the plane and hence be the normal vector of the plane.

If and , then we define as

Hence,

The vector equation of a plane is given by, , where is the position vector of a point on plane and is the normal vector.

If the co-ordinates of a point A≡ (x₁, y₁, z₁), then the position vector of A() is

Since B() lies on plane therefore ,

Hence, the vector equation is or or .

The cartesian equation will be or 5x+2y-3z=17

The new plane is parallel to the original plane and passes through point D≡(4,3,1).

Since both planes are parallel, therefore their normal vectors are same. Also, it passes through point D() therefore .

Hence, the vector equation is or or .

Or

Given that the plane contains line and point A≡(-1,3,-4).

Hence, the point and the vector lies on the plane.

If the co-ordinates of a point A≡ (x₁, y₁, z₁), then the position vector of A() is

Hence, the position vector of point A() is

If the position vector of point A() and B()is given then the vector .

Hence,

We have to find out the normal vector to plane. Since both and lies on plane, will be perpendicular to the plane and hence be the normal vector of the plane.

If and , then we define as

Hence,

The vector equation of a plane is given by, , where is the position vector of a point on plane and is the normal vector.

Hence, the vector equation is or or .

We have to figure out the intersection point of a line passing through point (2,1,4) and is perpendicular to the plane . Since, the line is perpendicular to plane, therefore it is parallel to it’s normal vector.

If a point on a line and a vector parallel to it is known, then the equation of line is .

Hence, the equation of the required line, passing through point and parallel to , is

For intersection point, substitute in the equation of the plane , hence

or λ-2+1+λ+4+λ=0

Solving, we get λ=-1

Hence, the point is or (3,0,3)

The distance between point (2,1,4) and (3,0,3) is . Hence the perpendicular distance from point (2,1,4) to plane is units.