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Q: Prove that if magnitude of vector A is constant then d by dt of vector A is perpendicular to vector A.?

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Try a vector approach.

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.

Draw a perpendicular to that line and extend the arms of the angle to meed the perpendicular drawn earlier. Check if the line is bisecting the perpendicular, if yes, then the line is a bisector of the angle. :)

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...

Actually The cross product of two vector is a VECTOR product. The direction of a vector product is found by the right hand rule. Consider two vectorsA and B,AxB= CWhere C is the Cross product of A and B, and by right hand rule its direction is opposite to that of BxA that isBxA=-C

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The unit vector is the ratio of the vector and its magnitude, thus : R/r = (Ix + Jy + Kz)/r where r= Sqroot(x^2 + y^2 + z^2). Units of the vector and the magnitude are the same thus divide out and the unit vector is dimensionless.

You don't need to prove much - just look at the definition of a vector. A vector includes a magnitude (in this case the force), and a direction. Since weight (or "the force of gravity") is directed to a certain direction, namely downward, you can consider it a vector.You don't need to prove much - just look at the definition of a vector. A vector includes a magnitude (in this case the force), and a direction. Since weight (or "the force of gravity") is directed to a certain direction, namely downward, you can consider it a vector.You don't need to prove much - just look at the definition of a vector. A vector includes a magnitude (in this case the force), and a direction. Since weight (or "the force of gravity") is directed to a certain direction, namely downward, you can consider it a vector.You don't need to prove much - just look at the definition of a vector. A vector includes a magnitude (in this case the force), and a direction. Since weight (or "the force of gravity") is directed to a certain direction, namely downward, you can consider it a vector.

A dot A = A2 do a derivative of both sides derivative (A) dot A + A dot derivative(A) =0 2(derivative (A) dot A)=0 (derivative (A) dot A)=0 A * derivative (A) * cos (theta) =0 => theta =90 A and derivative (A) are perpendicular

Suppose A is a vector with real components. A can be written as <f(t), g(t), h(t)>. Since the magnitude of A is constant we have f(t)*f(t) + g(t)*g(t) + h(t)*h(t) = c, where c is a non-negative real number. Take derivative of both sides of equation we get 2*f(t)*df(t)/dt + 2*g(t)*dg(t)/dt + 2*h(t)*dh(t)/dt = 0. Divide both sides by 2, we get f(t)*df(t)/dt + g(t)*dg(t)/dt + h(t)*dh(t)/dt = 0. Thus the dot product of A and its derivative is 0. This implies the angle between A and its derivative is Pi/2. Hence they are perpendicular.

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.

No. But then can you prove that you do?

Try a vector approach.

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.

Draw a perpendicular to that line and extend the arms of the angle to meed the perpendicular drawn earlier. Check if the line is bisecting the perpendicular, if yes, then the line is a bisector of the angle. :)

no be quiet

There is nothing to prove there. The second has been DEFINED to be the unit of time in the SI.

Show that corresponding angles are congruent?