Find vectors of length 10 parallel to the unit vector u=(3/5, 4/5)?
The unit vector has length 1 (by definition) and the required vector has length 10, so you just multiply each coordinate by 10. The required vector is 10(3/5, 4/5) = (30/5, 40/5) = (6, 8).
Find v x w, where v and w are vectors of length 3 in xz-plane, and (theta)=pi/6?
V x W will be perpendicular to V and W. Since V and W are on the xz-plane, V x W will be a vector that is parallel to the y-axis. That is a vector whose x and z components are 0 or (0, y, 0), and y is a number. ||V x W|| = ||V|| || W|| sin(theta) ||V x W|| = 3 * 3 sin(pi/6) = 9/2 so we know the length of the cross product is 9/2. The vector is (0,9/2,0) or (0,-9/2,0) note: depends on how you draw the vectors, you can use the right hand rule to pick the according one
There is only one such vector since you can place vectors anywhere. Think of vector only as directions. Let the given line have the following vector equation [math]l : p + t \cdot v[/math], where [math]p[/math] is a vector, [math]v[/math] is unit vector and [math]t[/math] is a real parameter. Then the vector [math]n \cdot v[/math] satisfies the requirements (it has length [math]n[/math] and has the same direction as the line). If you are asking for segments, then any segment [math][q, q+n \cdot v] [/math] is ok, where [math]q[/math] is any vector (point).
Need help with parallel vector problem?
vector QP = P - Q = (1, -3) - (3, -4) = (-2, 1) (from the origin) to get the vector of length 2 you simply need a normalizing factor: v = sqrt(4/5) (-2,1) = (-4/sqrt(5), 2/sqrt(5)) = (-1.789, 0.8944) is the required vector. To get two vectors parallel to QP of length two you can simply add v to any point. For example, start at the point A = (3,0) and end on the point B = (1.2111, 0.8944) and AB is a vector of length 2 parallel to PQ.
Parallel vector Question?
There are two parts of this. Parallel will come first: If you want to determine whether lines are parallel, compare the individual components to each other. If there is a consistent multiple to change one to the other, then the two vectors are parallel to the other. For example: ‹2, -3, 6› is parallel to ‹-2, 3, -6› because: x: 2 × -1 = -2 y: -3 × -1 = 3 z: 6 × -1 = -6 In other words, V×-1 = A In technical terms, if one vector is a "scalar multiple" of another vector, it is parallel. This means that all of the components must multiply by the same value to equal the resultant vector. However, this wants the "unit vector". You could probably realize that the two that are small numbers are the unit vectors of answers A and B. So the one the same signs as A (which would be parallel) would be the correct answer: C is the correct answer. But that's not how it is supposed to be done, so getting there goes like this: First, get the length ("norm") of the vector, which can be found by getting the square root of the sum of the squares of the components of the vector ‹2, -3, 6› : ||V|| = √( a² + b² + c² ) = √[ (2)² + (-3)² + (6)² ] = √( 4 + 9 + 36 ) = √( 49 ) = 7 The unit vector can be found be dividing the original vector by this value (or multiplying by the reciprocal as it is often written) ‹2, -3, 6› / 7 = 1/7 · ‹2, -3, 6› = ‹ 2/7, -3/7, 6/7› ≈ ‹ 0.29, -0.43, 0.86› From this, we can a parallel unit vector. The numbers of the last two are the same, which indicates that they would be unit vectors (while the first are clearly too long as they contain all non-zero integers). Since the numbers are all the same, you don't have to see if either is a multiple of it (which wouldn't be the case with unit vectors). It's just a matter of finding the one that matches the signs (or the opposite signs) of the original vector. -1 · ‹ 0.29, -0.43, 0.86› = ‹ -0.29, 0.43, -0.86› Which is, of course, option C (as deduced earlier). The signs are the opposite, which means it points in the opposite direction, but it would still be parallel.
THE more mathematically rigorous method -there is an operation on vectors defined as a UxV where U and V are your vectors, this operation is called the cross product. lets define this is equivalent to U = (u1,u2,u3) this is equivalent to V = (v1,v2,v3)if you dont already know what these i,j,k are, then these are simply the x,y,z components of the vector respectively :)now their cross product is defined asso now if the 2 vectors are parallel you will get UxV =0 that will be a zero vector = (0,0,0)lets take two vectors (1,1,1) and (2,2,2) now if you calculate its cross product it comes out to be 0i +0j+0k so they are parallel Now the Easier way ;)let your vectors be U = (u1,u2,u3) and V = (v1,v2,v3)now if (u1/v1)=(u2/v2)=(u3/v3) then the vectors are parallel you can check it your self for the case (1,1,1) and (2,2,2) :)
A vector is made up 2 elements that are- its direction and its magnitude.2 vectors would be parallel if they point in the same direction. if it's pointing towards x axis, the parallel vector should also point towards x axis,ORif it's pointing towards the direction of 45 degrees from the x axis, the parallel vector should also point to the same direction.ANDa unit vector would be a vector, whose magnitude is 1.So to find a unit vector parallel to the given resultant vector, we just make its magnitude 1 and give it the same direction as described in the example below! Let's say that the resultant vector is 6i, which means the vector points towards only one direction. Now, a vector of unit magnitude, and parallel to this would clearly be i or 1i, because it has a magnitude of one and the direction is same as that of the vector mentioned above.We got the answer just by the formula you described yourselfR cap=6i(resultant)/|6|(magnitude)= iSame would go for even 2D or 3D vectors! Hope it helped!
Are the Vectors Parallel, Orthogonal or Neither?
Parallel means that the vectors are headed in the exact same direction. Orthogonal means they are at a 90 degree angle. Vectors have both magnitude and direction. In this case, your vectors have the same magnitude, but are neither paralle nor orthogonal. We use the pythagorean theorem to find the magnitude. So each of these vectors has a magnitude of 5. But comparing the "slopes" of these vectors to compare direction we find v has a slope of -4/3, meanwhile w has a slope of -3/4. If these were parallel, then they would be the same. If they were orthogonal, then they would have a product of -1. However, these two multiply out to positive 1. So you answer is that v and w are neither parallel nor orthogonal.
[math]\vec{A}\times \vec{B}=|\vec{A}||\vec{B}|\sin\theta\cdot \hat{a}[/math]or,[math]|\vec{A}\times \vec{B}|=|\vec{A}||\vec{B}|\sin\theta[/math]In case of two parallel vectors, [math]\theta =n\pi[/math] where [math]n\in\Z[/math]which gives [math]\vec{A}\times\vec{B}=\vec{0}[/math]Done