Mastering physics which of the following vectors is equal to

Scalar products are used to define work and energy relations. For example, the work that a force a vector performs on an object while causing its displacement a vector is defined as a scalar product of the force vector with the displacement vector.

A quite different kind of multiplication is a vector multiplication of vectors. Taking a vector product of two vectors returns as a result a vector, as its name suggests. Vector products are used to define other derived vector quantities.

For example, in describing rotations, a vector quantity called torque is defined as a vector product of an applied force a vector and its distance from pivot to force a vector. It is important to distinguish between these two kinds of vector multiplications because the scalar product is a scalar quantity and a vector product is a vector quantity.

The scalar product is also called the dot product because of the dot notation that indicates it. The scalar product of a vector with itself is the square of its magnitude:. Figure 2. Substituting these values into Figure gives the scalar product. Show Answer. In the Cartesian coordinate system, scalar products of the unit vector of an axis with other unit vectors of axes always vanish because these unit vectors are orthogonal:.

For unit vectors of the axes, Figure gives the following identities:. We can use the commutative and distributive laws to derive various relations for vectors, such as expressing the dot product of two vectors in terms of their scalar components. When the vectors in Figure are given in their vector component forms,.

Since scalar products of two different unit vectors of axes give zero, and scalar products of unit vectors with themselves give one see Figure and Figure , there are only three nonzero terms in this expression.

Thus, the scalar product simplifies to. We can use Figure for the scalar product in terms of scalar components of vectors to find the angle between two vectors. Three dogs are pulling on a stick in different directions, as shown in Figure. Computing the scalar product of these vectors and their magnitudes, and substituting into Figure gives the angle of interest.

Notice that when vectors are given in terms of the unit vectors of axes, we can find the angle between them without knowing the specifics about the geographic directions the unit vectors represent. How much work is done by the first dog and by the second dog in Figure on the displacement in Figure? The magnitude of the vector product is defined as. The anticommutative property means the vector product reverses the sign when the order of multiplication is reversed:.

The corkscrew right-hand rule is a common mnemonic used to determine the direction of the vector product. The direction of the cross product is given by the progression of the corkscrew. The mechanical advantage that a familiar tool called a wrench provides Figure depends on magnitude F of the applied force, on its direction with respect to the wrench handle, and on how far from the nut this force is applied. To loosen a rusty nut, a Find the magnitude and direction of the torque applied to the nut.

The magnitude of this torque is. Physically, it means the wrench is most effective—giving us the best mechanical advantage—when we apply the force perpendicular to the wrench handle. In this way, we obtain the solution without reference to the corkscrew rule. Similar to the dot product Figure , the cross product has the following distributive property:.

The distributive property is applied frequently when vectors are expressed in their component forms, in terms of unit vectors of Cartesian axes.

We can repeat similar reasoning for the remaining pairs of unit vectors. The results of these multiplications are. The cross product of two different unit vectors is always a third unit vector. When two unit vectors in the cross product appear in the cyclic order, the result of such a multiplication is the remaining unit vector, as illustrated in Figure b.

When unit vectors in the cross product appear in a different order, the result is a unit vector that is antiparallel to the remaining unit vector i. In practice, when the task is to find cross products of vectors that are given in vector component form, this rule for the cross-multiplication of unit vectors is very useful.

These products have the positive sign. These products have the negative sign. We can use the distributive property Figure , the anticommutative property Figure , and the results in Figure and Figure for unit vectors to perform the following algebra:. When performing algebraic operations involving the cross product, be very careful about keeping the correct order of multiplication because the cross product is anticommutative.

The last two steps that we still have to do to complete our task are, first, grouping the terms that contain a common unit vector and, second, factoring. In this way we obtain the following very useful expression for the computation of the cross product:.

In this expression, the scalar components of the cross-product vector are. When finding the cross product, in practice, we can use either Figure or Figure , depending on which one of them seems to be less complex computationally.

They both lead to the same final result. One way to make sure if the final result is correct is to use them both. When moving in a magnetic field, some particles may experience a magnetic force.

To compute the vector product we can either use Figure or compute the product directly, whichever way is simpler. Hence, the magnetic force vector is perpendicular to the magnetic field vector. We could have saved some time if we had computed the scalar product earlier. Even without actually computing the scalar product, we can predict that the magnetic force vector must always be perpendicular to the magnetic field vector because of the way this vector is constructed.

The dot product is a scalar; the cross product is a vector. Later chapters use the terms dot product and scalar product interchangeably. Similarly, the terms cross product and vector product are used interchangeably.

How can you correct them? If the cross product of two vectors vanishes, what can you say about their directions? If the dot product of two vectors vanishes, what can you say about their directions? What is the dot product of a vector with the cross product that this vector has with another vector?

Why or why not? You fly [latex] For communication sake, we will refer to the first displacement as A and the second displacement as B. Since the magnitude and direction of the resultant is known, the x- and y-components can be determined using trigonometric functions.

Since the angle of degrees is expressed as a counterclockwise angle of rotation with due East, it can be used as the Theta in the equation. Whatever the magnitude and direction of B is, it must add on to vector A in order to give a southward displacement of This could be expressed by mathematical equations as.

By substituting these two values into the above equations, the values for the x- and y-components of the unknown displacement can be determined:. Knowing the B x and B y components will allow us to determine the magnitude and direction of B. Another diagram would help in visualizing the situation. The magnitude of B can be found using the Pythagorean theorem. B is the hypotenuse of a right triangle with legs of The sum of the squares of the sides is equal to the square of the hypotenuse.

The direction of B is close to degrees. As shown in the diagram, it is less than degrees by an amount equal to theta. The angle theta can be determined using the tangent function and the length of the two sides of the right triangle.

A boat heads straight across a river which is For the following two combinations of boat velocities and current velocities, determine the resultant velocity, the time required to cross the river, and the distance traveled downstream. Resultant Vel. The two velocity vectors boat and river are directed perpendicular to each other. They can be added using the Pythagorean theorem. The direction is found using the tangent function; it is expressed as the counterclockwise angle of rotation from due East.

The time to cross the river is dependent upon the width of the river and the boat velocity. And the distance downstream is dependent upon the time that the boat is moving and the speed at which it moves downstream - the river velocity. The solutions to all five of these projectile problems involve the use of kinematic equations and an appropriate problem-solving strategy.

The kinematic equations and their use in projectile problems are listed and discussed elsewhere. The basic idea of the strategy is to identify three kinematic variables for either the horizontal motion or for the vertical motion. Once , three quantities in one direction is known, all other quantities in that direction can be found or the time of flight can be found. Often, the time is then used with kinematic quantities for the second dimension in order to determine all other unknown quantities for that dimension.

When these three knowns are combined with the other given knowns the following answers are obtained:.

Use y, v iy , and a y to calculate t; then use t, v ix and a x to calculate x. Use t, v ix and a x to calculate x; and use t, v iy , and a y to calculate y. Use y, v iy , and a y to calculate t; then use t, x and a x to calculate v ix. Use t, x and a x to calculate v ix ; and use t, v iy , and a y to calculate y. The launch velocity and angle is given for three different projectiles. Use trigonometric functions to resolve the velocity vectors into horizontal and vertical velocity components.

Then use kinematic equations to determine the time that the projectile is in the air, the height to which it travels when it is at its peak , and the horizontal distance that it travels.

The solutions to all three of these non-horizontally launched projectile problems involve the use of kinematic equations and an appropriate problem-solving strategy. The values of v ix and v iy can be determined using trigonometric functions:. Once v ix and v iy are known, the other unknowns can be calculated. The time up to the peak t up can be determined using the equation. Once t up is known, the t total time to travel the entire trajectory -both up and down can be determined by doubling the t up.

The horizontal displacement of the projectile x can be computed in the usual way using the equation. Finally, the y at the peak i. The t value in the equation is t up because the peak height is reached when the projectile has traveled for one-half of its total time; t up is that time. This method will yield the answers given in the table above. If a projectile is launched horizontally with a speed of Determine the horizontal displacement of the projectile.

This horizontally-launched projectile problem can be and should be solved in the same manner as the solution to 60 above. While 60 is broken down for you into nice steps, this problem is not so user-friendly. It is strongly recommended that you begin by listing known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.

Since three pieces of y-information are now known, a y-equation can be employed to find the time. Plugging and chugging the above values into this equation yields a time of 2. Now the t value can be combined with the v ix and a x value and used in an x-equation. More examples and discussion of these types of projectile problems are discussed elsewhere.

A projectile is launched with an initial speed of This non-horizontally-launched projectile problem can be and should be solved in the same manner as the solution to 61 above. While 61 is broken down for you into nicely-structured steps, this problem is not so user-friendly.

It is strongly recommended that you begin by resolving the initial velocity and angle into initial velocity components using the equations:. Once done, list the known values for each of the variables in the kinematic equations.

One useful equation is. These two solutions to the equation indicate that the time is 0 s when the vertical displacement y is 0 m. The latter of the two solutions can be used to determine the horizontal displacement x. Use the equation:. Plugging a chugging the above values into this equation yields the answer of Finding the vertical displacement at the peak y peak demands using the original y equation with a time of 1.

This time corresponds to the time for one-half of the trajectory - the time at which the projectile will be at its highest or peak position. Substituting the v iy , a y and t values into the equation.

A projectile is launched horizontally from the top of a Determine the magnitude of the launch velocity. The best means of starting this problem is to list the known values for each of the variables in the kinematic equations. A full parabola which follows the above function would have to locations where the y coordinate is One would be "forward in time" at 3. Of course, the positive answer is the one which we need; it can can be used to determine the initial horizontal velocity v ix.

Plugging and chugging the above values into this equation yields the answer of 5. Two physics students stand on the top of their 3. The balloon is launched upward at a speed of The balloon lands in a retention pond whose surface is 2.

Determine the horizontal distance from launch location to landing location. This is a non-horizontally-launched projectile problem in which the initial velocity and launch angle are given. Three initial steps are always wisely taken before starting such a problem. First, determine the initial velocity components v ix and v iy using trigonometric functions.

Second, construct a diagram of the physical situation. And third, organize known and unknown information in an "x-y table. Quickly barging into a solution before giving the problem some pre-analysis often leads to a wasting of much time and ultimately a lot of confusion. The initial velocity and angle can be resolved into initial velocity components using the equations:.

Now the known values for each of the variables in the kinematic equations is listed in a table using a column for known horizontal information and a column for known vertical information.

A full parabola which follows the above function would have two locations where the y coordinate is One location would be "forward in time" at 6. Of course, the positive answer is the one which we need; it can can be used to determine the horizontal displacement x. Plugging and chugging the above values into this equation yields the answer of m. Those boys had better be careful.

A place kicker kicks a football from The kick leaves the ground with a speed of The goal posts are 3. Assume that the football hits the horizontal crossbar of the posts and bounces through. Given: 1. The height of the goal posts can be subtracted from this value to determine the amount of clearance.

As is the case in all non-horizontally-launched projectile problems, it should be begun by resolving the initial velocity and angle into initial velocity components using the equations:.

Since three pieces of x-information are now known, an x-equation can be employed to find the time for the football to travel the horizontal distance to the goal posts. The time can now be combined with a y-equations to find the vertical displacement i. When the ball has traveled a horizontal distance of The task will involve first finding the time for the ball to rise to its peak and then fall back down to a height of 3.

Then the horizontal displacement can be calculated for this time. The final answer will then need to be converted to yards. The same v ix and v iy values can be used. Given the new context of the problem, the value for y is now known and x is unknown. The information can be organized in the usual x- and y- table. The first solution corresponds to the first point along the parabola during the rise of the football when the football is at a height of 3.

The second answer can be used to determine the horizontal displacement x of the football. This x value can be converted to feet by multiplying by the 3. The kicker can kick as Field goal distances are are measured from the goal line to the line of scrimmage. An airplane starts at Point A and flies The plane then flies km at degrees to Point C. Finally, the plane flies km at 29 degrees to Point D.

Determine the resulting displacement magnitude and direction from Points A to D. Like most vector addition problems, this problem is best begun by the construction of a rough sketch of the physical situation.

This sketch is shown below. There are three separate displacements taken by the airplane to result in a single displacement from point A to point D. The actual solution is best performed using trigonometric functions to determine the x- and y-components of each displacement vector.