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
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 \( \
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 of a plane containingP_0=(1,2,3)and perpendicular ton=[1,-1,2]. Solution:(x-1)-(y-2)+2(z-3)=0\Rightarrow x-y+2z=5 3.3 The line of intersection of two planes Example Find a vector tangent to the line of intersection of the planes ...
Sometimes it is useful to replace the short vector forms of the equation of a plane with a form in which the vectors are replaced with their components: x = 〈x, Y, z〉 and n = 〈xn, yn, zn〉. Thus, expanding Eq. (5.11) gives the following equation....
b问:第一步,平面p和平面q互相平行,通过这个条件可以得到,两个平面的法线是相同的,所以平面q的表达式可写成2x-4y-z=d 第二步,平面经过点(1,-2,5),所以这个点满足上述表达式,代入其中 2×1-4×(-2)-5=d,得到d=5 结论The plane q has equation 2x-4y-z=5 ...
A vector equation is a mathematical expression that represents a straight line or a plane using a combination of a fixed point and a direction vector. AI generated definition based on: Mathematical Formulas for Industrial and Mechanical Engineering, 2014 ...
Vector Equation Of A Plane The vector equation of a plane which passes through a point {eq}P(x,y,z) {/eq} and containing the vectors {eq}a {/eq} and {eq}b {/eq} is given by {eq}R(u,v)=(x\mathbf i+y \mathbf j+z \mathbf k)+\mathbf a u+ \mathbf b ...
The distance from that point to (0,0,0) is (4/81)A2+(4/81)A2+(1/81)A2=(1/3)A. In order that that be "5" we must have A= 15. The equation of the plane is 2x- 2y+ z= 15. In vector notation that is (2,−2,1)⋅(x,y,z)−15=0 H0...