Find a vector i normal to the plane 2x + y + 2z = 4
1 answer:
Answer: <em>i</em> =2i+1j+2k
Step-by-step explanation:
The parametric equation of a plane is Ax+By+Cz=D and by definition the normal vector of the plane is n=Ai+Bj+Ck, where i,j,k are the canonic vectors
Then, in this case a normal vector to the given plane is i=2i+1j+2k
To prove that a vector defined as n=Ai+Bj+Ck is normal to the plane, let V1 and V2 be two vectors that are in the plane
If V1 ∈ Ax+By+Cz=D then Aa1+Bb1+Cc1=D
If V2 ∈ Ax+By+Cz=D then Aa2+Bb2+Cc2=D
The vector V1V2=V2-V1=(a2-a1)i+(b2-b1)j+(c2-c1)k
Two vectors are normal or perpendicular if the dot product is zero.
So, n is a normal vector to the plane.
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Answer:
x = 15
Step-by-step explanation:
The easiest way to solve this is to realise that a triangle takes up half the area of a rectangle of the same width and height.
We are told that the width of the triangle is 10, and that the line of length 10 is perpendicular to the longest side of the triangle. Because of that we know that x can be multiplied by ten to get the area of the rectangle that is twice the area of the triangle.
We are also told that the triangle's area is 75 units. With all of that put together, we can say: