![]() So 9+4+36, now this is 13+36 which is 49 and root 49 is 7. The idea is the same: multiply corresponding elements of both column matrices, then add up all the products. What's the magnitude of the vector 3, 2, 6 it's the square root of the sum of the squares of the components. The dot product is defined for 3D column matrices. And that's negative 5 plus 30 25, how does subtraction work? Again exactly the same as with 2 dimensions you subtract like components 1-2 negative 1, 2 minus negative 1 is 3, 1-1 0. This is a useful result when we want to check. How about the dot product of these 2 guys it's going to be 1 times 3, 3 plus negative 4 times 2 negative 8 plus 5 times 6 30. If 2 vectors act perpendicular to each other, the dot product (ie scalar product) of the 2 vectors has value zero. So I'm going to do the scalar multiplication first 2 times 4 negative 1, 2 is 8 negative 2, 4 and I'll add that to 3, 8 negative 2 so I get 3+8 is 11, 8 plus negative 2 is 6 and negative 2 plus 4 is 2. Let's do an example that exercises these rules, so part a asks me to simplify 3, 8 negative 2 plus 2 times 4 negative 1 and 2. So u1 squared plus u2 squared plus u3 squared okay. The dot product works exactly the same way too it's going to be the product of like components so u1v1+u2v2+u3v3, u1v1+u2v2+u3v3 and finally the magnitude of a vector u is just the square root of the sum of the squares of the components. What about scalar multiples? Suppose k is some real number and you're multiplying it by vector u, well same as before it's k times u1, k times u2, k times u3, k times each of the components. Example: Calculate the dot product of vectors a and b: 10 × 13 × cos(59.5) 10 × 13 × 0.5075. ![]() Declaring vector1 and initializing x,y,z values Vector3D vector1 new Vector3D(20, 30, 40) // Declaring. In 3 dimensions you have 3 components u1, u2 and u3, v1, v2 and v3 so how do you add vectors? Just the same as in 2 dimensions you add them component wise so the sum will be u1+v1, u2+v2, u3+v3. Calculates the Dot Product of two Vectors. So how do vector operations work in 3 dimensions? It turns out it's almost exactly the same way as two dimensions only you have a new component, now let's take 2 vectors u and v.
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