write the equation for the planethe plane through the point P(7,-4,-5)and normal to n =4i 8j 2k什么叫nornmal to n 相关知识点: 试题来源: 解析 normal to n =4i 8j 2k 以n =4i 8j 2k为法向量.也就是垂直于n =4i 8j 2k
We find a vector normal to the plane and use it with one of the points (it does not matter which) to write an equation for the plane.The cross productis normal to the plane. We substitute the components of this vector and the coordinates of into the component form of the equation to...
解析 Since the two planes are parallel, they will have the same normal vectors. A normal vector for the plane z=x+y orx+y-z=0 is , and an equation of the desired plane is 1(x-2)+1(y-4)-1(z-6)=0 or x+y-z=0 (the same plane!)....
(1)with normal vector (bmatrix)2 -1 3(bmatrix) and through (-1,2,4) (2)perpendicular to the line through (2, 3, 1) and (5, 7, 2) and through (3)perpendicular to the line connecting (1,4, 2) and (4, 1, -4) and containing such that (AP):(PB)=1:2 ...
The required normal vector to the plane is,n=(PQ)* (PR)=(vmatrix) i&j&k -1&1&0 -1&0&1(vmatrix) = i+j+kSo, the plane equation is1(x-1)+1(y-0)+1(z-0)=0x+y+z-1=0x+y+z=1反馈 收藏
{0}} \right)+q\left( y-{{y}_{0}} \right)+r\left( z-{{z}_{0}} \right)=0 {/eq} because the normal to the plane will be proportional to {eq}\left( p,q,r \right) {/eq} and the plane passes through the point {eq}\left( {{x}...
EXAMPLE 2 Find the equation of the tangent plane and the normal line to the surface x2 + y2 + 2z2 = 23 at (1, 2, 3). 相关知识点: 试题来源: 解析 Theorem A Tangent Planes For the surface F(x, y, z) = k, the equation of the tangent plane at (xo, yo, zo) is VF(xo. ...
When there are three points given in a plane, then first of all find the required vectors, take their cross product. This vector will be the normal vector to the plane {eq}n=\langle a,b,c \rangle {/eq}. Then equation of plane pas...
The plane is also parallel to the vector connecting the two points (-3, 1, 1) and ( 8/3, 7/3, 0), which is the vector (17/3, 4/3, -1). The plane is parallel to both vectors (-1, 2, 1) and (17...
Let the required equation of the plane be ax+by+cz+d=0. The d.r.'s of the plane are a,b,c. The d.r's of the x-axis are 1,0,0. Normal of the required plane is perpendicular to the x-axis. therefore (axx1)+(b xx0)+(c xx 0)= 0 rArr a=0. Hence, th