0
Prove that if magnitude of vector A is constant then d by dt of vector A is perpendicular to vector A.?
Wiki User
∙ 10y ago
That is not even true!
Wiki User
∙ 10y ago
Add your answer:
Earn +20 pts
How can you prove parallelogram inequality?
Try a vector approach.
How do you prove using a diagram that the magnitude of the sum of two vectors is less than or equal to the sum of the magnitude of the two vectors?
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.
What do you need to do to prove a line is an angle bisector of an angle?
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. :)
If a and b are two vectors and the vector product of vector a and vector b is equal to a minus b then prove that a equals b?
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...
If the commutative law not hold in cross product then prove it?
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
Prove that the unit vector is dimenssionless?
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.
How do you prove that weight is a vector quantity?
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.
Prove that a constant vector always has a perpendicular derivative?
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
Prove that if vector A has constant magnitude then its derivative is perpendicular to vector A?
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.
Prove that two vectors must have equal magnitude if their sum is perpendicular to their difference?
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.
Does null vector have real existence?
No. But then can you prove that you do?
How can you prove parallelogram inequality?
Try a vector approach.
How do you prove using a diagram that the magnitude of the sum of two vectors is less than or equal to the sum of the magnitude of the two vectors?
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.
What do you need to do to prove a line is an angle bisector of an angle?
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. :)
How do you prove that the molar heat capacity at the constant volume?
no be quiet
How do you prove that the unit of time constant is seconds?
There is nothing to prove there. The second has been DEFINED to be the unit of time in the SI.
How can you prove that a constructed line is parallel to a given line if the transversal in not perpendicular?
Show that corresponding angles are congruent?
