Find the equation of the 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...
Step 1: Identify the normal vectorThe equation of the plane can be expressed in the form →r⋅→n=d, where →n is the normal vector to the plane. From the given equation, we can identify the normal vector →n:→n=2^i+^j+2^k Step 2: Calculate the magnitude of the normal vecto...
Normal Vectors Lesson Summary Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Recommended Lessons and Courses for You Related Lessons Related Courses Cross Product of Two Vectors | Formula, Equation & Examples Tangent Plane to a Surface | ...
To find its equation, we need to find a normal vector to this plane, that is a vector which is perpendicular to this plane. Answer and Explanation: The plane passes through three points A(3,−1,2),B(6,3,5), and C(−1,−3,−3). These three points g...
To find the vector equation of the plane and its Cartesian form, we can follow these steps:Step 1: Identify the normal vector and distance from the origin The normal vector \( \mathbf{n} \) is given as \( 2 \hat{i} - 3 \hat{j}
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
Consider the plane which passes through the three points: {eq}(-7, -3, 10), (-10, -7, 13), {/eq} and {eq}(-10, -6, 15) {/eq}. Find the vector normal to this plane which has the form: {eq}(11...
The equation of a plane with normal vector {eq}\langle a,b,c \rangle {/eq} and passing through {eq}(x_1,y_1,z_1) {/eq} is {eq}a(x-x_1)+b(y-y_1)+c(z-z_1)=0 {/eq} Answer and Explanation: The normal vector of the r...
The equation of a line is linear in two variables, usually x and y, that is satisfied by points of the line. The equation of line is usually found by the point-slope form y - y_1 = m (x - x_1), where m is the slope and (x_1, y_1) is a point on the line.
Once I have the equation of the curve i have to discretize it to small parts and find the radius of the curve. I have another program that has a function which gives me an output based on the radius of the curve. what would be ideal is I find the curv...