作为垂直于它的单位向量。解决方案：
We know, the vector equation of a plane normal to unit vector
Here, d = 3 units ,we get
解决方案：
We know, the vector equation of a plane normal to unit vector
Here, d = 5 units
and
Hence the required equation is,
解决方案：
Dividing the equation by
Since Vector equation of a plane with distance d and normal to the unit vector n is given by
Comparing (1) and (2), we get
Distance from origin = 2 units
Direction cosine of normal to plane =
到正常形式，因此找到从原点到平面的垂直长度。解决方案：
Multiplying both sides by –1, we get
Dividing (1) by 3 on both sides,
Since vector equation of a plane with distance d and normal to the unit vector n is given by
Comparing (1) and (2), we get
d = 2
Length of normal = 2 units.
解决方案：
Multiplying both sides by –1, we get
Dividing (1) by 7 on both sides,
Hence normal form of the equation is
解决方案：
Normal vector =
So, Normal unit vector
Since vector equation of a plane with distance d and normal to the unit vector n is given by
or,
解决方案：
Dividing (1) by
Since vector equation of a plane with distance d and normal to the unit vector n is given by
Thus, normal unit vector =
从原点和法线到与坐标轴相同倾斜的单位。解决方案：
Since vector equation of a plane with distance d and normal to the unit vector n is given by
d =
Let
Since
We know, l2 + m2 + n2 = 1
or,
Now,
Vector equation of the required plane is
or,
or, x + y + z = 9.
解决方案：
Vector equation of a plane is given by
We have,
and,
Putting
Dividing (1) by
Hence, vector equation of plane is
and, cartesian form is x – y + 3z – 2 = 0.
从原点和它从原点的法向量是
.此外，找到它的笛卡尔形式。解决方案：
Since vector equation of a plane with distance d and normal to the unit vector n is given by
Since,
Unit vector normal to the plane =
or,
Vector equation becomes,
Cartesian equation is 2x – 3y + 4z = 6.
解决方案：
2x – 3y + 4z – 6 = 0
or, 2x – 3y + 4z = 6
Vector equation becomes,
or,
Dividing (1) by
Hence, the perpendicular distance of the origin from the plane is
