Find a vector perpendicular to the plane $2 \cdot x - y + 2 z = 5$ $2*x-y+2z=5$ ?

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Answer 1 · 2016-09-17T04:02:23

The Rqd. Vector is $(2, - 1, 2)$ $(2,-1,2)$ .

If the eqn. of a plane $π$ $pi$ is $π : a x + b y + c z + d = 0$ $pi : ax+by+cz+d=0$ , then, the the

vector $⊥$ $bot$ to $π$ $pi$ , called the normal vector of $π$ $pi$ is the

vector $(a, b, c)$ $(a,b,c)$ .

In our case, it is the vector $(2, - 1, 2)$ $(2,-1,2)$ .

Find a vector perpendicular to the plane 2*x-y+2z=52⋅x−y+2z=52*x-y+2z=5 ?