Given the equation of a plane, {eq}\displaystyle x - 2y + z + 1 = 0 {/eq} We need to find the normal vector to the plane. The normal form of a... See full answer below.Become a member and unlock all Study Answers Start today. Try it now Create an account Ask ...
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...
Find a unit vector normal to the plane containing u = i + 2j + 2k and v = - i + 3j - k. Find a unit vector normal to the plane containing u=3i+2j-k and v=-i-2j-3k. Find the unit vector normal to the plane containing (1,-2,2), (0,3,-1) and ...
View Solution Find the vector equation of the line through the points(1,−2,1)and(0,−2,3). Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation...
The line through the point (1,0,6) and perpendicular to the plane x+y+z=5. 相关知识点: 试题来源: 解析 A line perpendicular to the given plane has the same direction as a normal vector to the plane. such as n=(1,3,1). So r_0= i+6 k, and we can take v= i+3 j+ k...
解析 A normal vector to the tangent plane (and the surface) at (0,0,1) is (1,0,-1). Then parametric equations for the normal line there are x=t, y=0, z=1-t, and symmetric equations are x=1-z, y=0.结果一 题目 Find equations of the normal line to the given surface at ...
Answer to: Find a vector normal to the surface h(x,y) = 3x^2-x^3 at x = 4, y = 2. By signing up, you'll get thousands of step-by-step solutions to...
解析 A normal vector to the tangent plane (and the surface) at (1,-2,1) is 8,4,-1. Then parametric equations for the normal line there are x=1+8t, y=-2+4t, z=1-t, and symmetric equations are (x-1)8= (y+2)4= (z-1)(-1)....
Find a unit vector perpendicular to the plane containing the vectors ... 03:43 Find the magnitude of vector vec a=(3 hat k+4 hat j)xx( hat i+ hat j-... 03:00 If vec a=4 hat i+3 hat j+ hat k\ a n d\ vec b= hat i-2 hat k\ t h e ... 02:51 If vec a=3 ha...
P(1, 1, 1), Q(2, 1, 3), R(3, -1, 1) 相关知识点: 试题来源: 解析 (b) Unit vector ⊥ to plane PQR ( (PQ)* (PR))/(| (PQ)* (PR)|)= (4i+4j-2k)/6= (2(2i+2j-k))/6= 2/3i+ 2/3j- 1/3kThus, the solution is ( 2/3i+ 2/3j- 1/3k)...