You could draw a circle [center at origin] with radius of (a + b), for the two magnitudes a and b. This represents the sum of the magnitudes. Then draw one of the vectors starting at the origin [suppose it's vector a], and then draw a circle centered at the endpoint of vector a, with a radius of b. Drawing a circle demonstrates how the second vector can point in any direction relative to the first vector. The distance from the origin to a point on this second circle is the magnitude of the resultant vector. Graphically this second circle will be entirely inside the first circle and touching it at just one point. Since it lies within the first circle, the distance from the origin to a point on that circle will be less than or equal to the radius of the first circle.
If the sum of their components in any two orthogonal directions is zero, the resultant is zero. Alternatively, show that the resultant of any two vectors has the same magnitude but opposite direction to the third.
If the vectors emanting from one corner of the rectangel are called a and b then. (a) + (b) = one diagonal (a) + (-b) = the other diagonal and |(a) + (b)| = |(a) + (-b)| (the absolute value of the diagonal's scalars are equal)
The question is not correct, because the product of any two vectors is just a number, while when you subtract to vectors the result is also a vector. So you can't compare two different things...
using ven diagram prove de morgans law
You prove that the two sides (not the bases) are equal in length. Or that the base angles are equal measure.
If the sum of their components in any two orthogonal directions is zero, the resultant is zero. Alternatively, show that the resultant of any two vectors has the same magnitude but opposite direction to the third.
Suppose the condition stated in this problem holds for the two vectors a and b. If the sum a+b is perpendicular to the difference a-b then the dot product of these two vectors is zero: (a + b) · (a - b) = 0 Use the distributive property of the dot product to expand the left side of this equation. We get: a · a - a · b + b · a - b · b But the dot product of a vector with itself gives the magnitude squared: a · a = a2 x + a2 y + a2 z = a2 (likewise b · b = b2) and the dot product is commutative: a · b = b · a. Using these facts, we then have a2 - a · b + a · b + b2 = 0 , which gives: a2 - b2 = 0 =) a2 = b2 Since the magnitude of a vector must be a positive number, this implies a = b and so vectors a and b have the same magnitude.
If the vectors emanting from one corner of the rectangel are called a and b then. (a) + (b) = one diagonal (a) + (-b) = the other diagonal and |(a) + (b)| = |(a) + (-b)| (the absolute value of the diagonal's scalars are equal)
The question is not correct, because the product of any two vectors is just a number, while when you subtract to vectors the result is also a vector. So you can't compare two different things...
using ven diagram prove de morgans law
You prove that the two sides (not the bases) are equal in length. Or that the base angles are equal measure.
That is not even true!
You cannot prove that because it's false
prove any two adjacent triangles as congruent
Once you prove that a diagram is congruent then you can say that all the parts are congruent.
It isn't equal, and any proof that they are equal is flawed.
No, but there is a way to prove that zero equals one.