解析 x+y=2. The direction vectors of the lines are (1,-1,2) and (-1,1,0), so a normal vector for the plane is(-1,1,0)* (1,-1,2)=(2,2,0) and it contains the point (2,0,2). Then an equation of the plane is 2(x-2)+2(y-0)+0(z-2)=0⇔ x+y=2....
百度试题 结果1 题目25. Find the equation of a plane containin gthepsints(0,-1,-1),(-4,4,4)and (4, 5, 1). Also show that (3, 9, 4) lies on that plane. 相关知识点: 试题来源: 解析 5x-7y+11z+4=0. 反馈 收藏
15Findtheparametricequationsforthelinethroughthe…
Answer to: Find the equation of the plane. The plane that passes through the point (-1, 2, 1) and contains the line of intersection of the planes x...
1.) Find the equation of the plane that contains both lines. Equation of a Plane: Recall that if we have a point {eq}(x_0,y_0,z_0) {/eq} in a plane and a normal to the plane {eq}\vec n = \left {/eq}...
【题目 】Solve the following problem(1)Find the equation of the plane through(-1, 0, 2), (0,-1, 1) and c(1,2,-1):(2)Find the equation of the line, in parametric form, which passes through the origin and is normal to the plane in (1).(3)Find the point where the line ...
解析 Given equation of plane is 3x+4y-5z= 0Equation of plane parallel to 3x+4y-5z= 0is given by3x+4y-5z+k= 0Since it passes through the point (1.2.3)3×1+4×2-5×3+k=03+8-15+k=0k-4substituent the value of the k we get3x+4y-5z+4=0 ...
Find the equation of plane containing the line x+ y - z = 0 = 2x – y + z and passing through the point (1, 2, 1)
Answer to: Find the equation of the plane containing the points (1,1,1) and (3,0,4) and the vector <1,1,-1>. By signing up, you'll get thousands of...
Find an equation of the plane containing(1,−1,3)and perpendicular to the vectorn=(2,1,4). Equation of the Plane : (x1,y1,z1) and is perpendicular to the vectorn=(a,b,c)is given by: a(x−x1)+b(y−y1)+c(z−z1)=0 ...