The vector equation of a plane is given by, (r-a).n=0, where a is the position vector of a point on plane and n is the normal vector.Hence, the vector equation is (r- i- j ).(-6 i+6 j+6 k)=0 or r.(-6 i+6 j+6 k)=0 or r.(- i+ j+ k)=0.We have to ...
Find the equation a plane containing the line vecr =t veca and perpend... 03:23 Prove that the line of intersection of x + 2y + 3z=0 and 3x + zy+ z=0 ... 04:07 Find the distance of the point (3,8,2) from the line vecr=hati + 3 hat... 05:30 Find the vector equation...
百度试题 结果1 题目(1) Find the vector equation of a plane which is at 42 unit distance from the origin and which is normal to the vector 2i+ j-2k. 相关知识点: 试题来源: 解析 ·(2i+j-2k)=126 反馈 收藏
Vector equations ares used to represent the equation of a line or a plane with the help of the variables x, y, z. The vector equation defines the placement of the line or a plane in the three-dimensional framework. The vector equation of a line is r = a + λb, and the vector ...
To find the vector equation of a plane passing through a given point and perpendicular to a specified vector, we can follow these steps:Step 1: Identify the given information We have: - A point \( A \) with position vector \( \
independent的vector做basis， 比如(2,3,4) - (-2,1,5) = (4, 2, -1), (-1, 1, 1).再随便找个起点，比如(-2, 1, 5)所以该平面就是(-2 +4a -b, 1 + 2a + b, 5 -a +b)另外，vector只有方向，没有起点，不存在神马vector of line共起点这一说 ...
Find a vector valued function whose graph is the surface given by x^{2} + y^{2} = a^{2} z^{2}. Also identify the type of surface. Find a unit vector perpendicular to the surface x2 + y2 + z2 = 3 at the point (1,1,1). Then derive the equation of the ...
Recall that (a) Find an equation of a plane through which is perpendicular to the (b) Find an equation of the plane which is perpendicular to the line at the point when t = 2 (c) Find the equation of the plane which is equidistant from the points Bonus. Parametric Plane Can we use...
119 -- 19:03 App 8.4 Vector and Parametric Equation of a Plane 36 -- 18:31 App 8.3 Vector Parametric Symmetric Equations of a Line in R³ 24 -- 14:15 App 6.1 Introduction to Vector 41 -- 11:13 App 13 Vector Equation & Vectors in R^2 222 -- 3:57 App P3 11.1 Vector...
In summary, the equation of a plane can be found by solving for the normal vector equation of a plane which is normal on the vector n=(2,-2,1), at a distance of 5 from (0,0,0). 1 Apr 7, 2008 #36 Physicsissuef 908 0 Hootenanny said: Note that...