Learn Vectors with Examples, Equations, and Practice Problems

A plane passes through the point (1,1,1) and is perpendicular to each of the planes
3x−2y+3z+6=0 and 6x−2y−3z−6=0. Find its equation. The problem is I don't have an idea of the concept. All I know is that the normal of first equation is (3,−2,3) and that of the second is (6,−2,−3).

Dangelo Rosario 2022-10-02

Proof of: $u_{j} \frac{\partial}{\partial x_{i}} u_{j} = \frac{1}{2} \frac{\partial}{\partial x_{i}} (u_{j}^{2})$
I'm reading a proof of:
$u \times ω = \nabla (\frac{u \cdot u}{2}) - u \cdot \nabla u$
In this proof, it says:
$u_{j} \frac{\partial}{\partial x_{i}} u_{j} = \frac{1}{2} \frac{\partial}{\partial x_{i}} (u_{j}^{2})$
It might be trivial, but I cannot get my head around this later equality. Does anyone know how to prove/motivate it (using Einstein notation)?

dripcima24 2022-10-02

I am trying to understand dot and cross products from a physics perspective. If a space curve r(t) satisfies the equation $r^{'} (t) \times r^{″} (t) = 0$ for all t, I understand that the derivative of r′(t) is parallel to r′(t), so the velocity vector r′(t) does not change direction. Thus, this curve moves along a line.
However, I'm not sure what $r^{'} (t) \cdot r^{″} (t) = 0$ means. My intuition is that it represents motion along a circle, or part of a circle, since the velocity and acceleration vectors should be perpendicular in that case.

Parker Pitts 2022-10-01

If M is a linear operator on $R^{3}$ with unique and real eigenvalues $λ_{1} < λ_{2} < λ_{3}$ , such that $\exists x \in R^{3} ∖ {0}$ , satisfying the condition $lim_{n \to \infty} | | M^{n} x | | = 0$ . What are the possible values of $λ_{1}$ ?

Jamarcus Lindsey 2022-10-01

Let's say I have a vector u, whose magnitude (L2-norm) is $| u |$ and whose direction is $\frac{u}{| u |}$ . I have another vector v, whose magnitude and direction are both different to that of u. So, $| v | \neq | u |$ and $\frac{v}{| v |} \neq \frac{u}{| u |}$
What I want to do, is create a new vector w whose magnitude is equal to that of u, but whose direction is equal to that of v. So this can be thought of as taking vector v, and changing its magnitude such that the magnitude is equal to $| u |$
This seems quite trivial, but I cannot work it out! Any help please?

Nathanial Levine 2022-10-01

I have three vectors: $\vec{u}$ , $\vec{v}$ and $\vec{w}$ and this equality: $\vec{u} \cdot \vec{v} = \vec{u} \cdot \vec{w}$ . Then the question is if $\vec{v} = \vec{w}$ . At first i tried to write the equality representing components:
$\vec{u} \cdot \vec{v} = u_{1} \cdot v_{1} + \dots + u_{n} \cdot v_{n}$
$\vec{u} \cdot \vec{w} = u_{1} \cdot w_{1} + \dots + u_{n} \cdot w_{n}$
Then:
$u_{1} \cdot v_{1} + \dots + u_{n} \cdot v_{n} = u_{1} \cdot w_{1} + \dots + u_{n} \cdot w_{n} .$
But I got stuck right there; I thought if $c_{i}, i \in [1, n]$ could be canceled from both terms, but don't know if that is possible somehow.

Dangelo Rosario 2022-10-01

Given that u is a d x 1 vector such that ||u|| = 1 and for all d x 1 vector x, how can we reason through the equation ? $x^{T} (u u^{T}) x = (x^{T} u) (u^{T} x) = (⟨ x, u ⟩) (⟨ x, u ⟩) = ((⟨ x, u ⟩))^{2} \geq 0$
In this case, x is a scalar(?). and $(u u^{T})$ gives us 1(?) which mean $\sum_{1}^{d}$ $x_{1} \cdot x_{1}$ , Then since ||u|| is 1 given. what can we say about each $x_{1}$ if x is a scalar, can we even transpose it ?

gsragator9 2022-09-30

Find the value of k for which the plane $k x + 4 y + 2 z - 6 = 0$ is parallel to the line $\frac{x - 3}{5} = \frac{y}{1} = \frac{z}{- 3}$
So based on this I know the normal to the plane is $(k, 4, 2)$ . I also know the direction vector of the line is $(5, 1, - 3)$
I need to find a vector that is perpendicular to the normal, and then somehow use that with the given direction vector to find k, but I'm not quite sure how.

Deanna Gregory 2022-09-30

In a longer derivation I ran into the following quantity:
$\nabla [\nabla \cdot (r_{0} e^{i k \cdot r})]$
( i.e., the gradient of the divergence ) where k is a vector of constants and r is a position vector.
Can someone help explaining how to calculate this? I am hoping it gives:
$\nabla [\nabla \cdot (r_{0} e^{i k \cdot r})] = - k (k \cdot r_{0})$

abkapseln87 2022-09-29

Suppose b is an n-dimensional column vector that can be decomposed as $b = b_{1} + b_{2}$ . In this decomposition, $b_{1}$ and $b_{2}$ hold the non-zero components of $b$ respectively (e.g. if $b$ = ${[\begin{matrix} 1 & 2 & 3 \end{matrix}]}^{⊤}$ , then this decomposition can be $b_{1}$ = ${[\begin{matrix} 1 & 0 & 3 \end{matrix}]}^{⊤}$ and $b 2$ = ${[\begin{matrix} 0 & 2 & 0 \end{matrix}]}^{⊤}$ ).
Suppose A is a real matrix of order n and let $C$ =[ $b$ , $A b$ , ... , $A^{n - 1} b$ ], $C_{1}$ =[ $b_{1}$ , $A b_{1}$ , ... , $A^{n - 1} b_{1}$ ] and $C_{2}$ =[ $b_{2}$ , $A b_{2}$ , ... , $A^{n - 1} b_{2}$ ]. If we have rank([ $C_{1}$ , $C_{2}$ ])=n, can we prove rank(C)=n?

beninar6u 2022-09-29

Given the operation in $R^{2}$ :
$(x_{1}, y_{1}) + (x_{2}, y_{2}) = (x_{1} x_{2}, y_{1} y_{2})$
I would like to find whether this is a vector space in $R$ . Looking at the Additive Zero Axiom, we get:
$(x_{1}, y_{1}) + 0 = (x_{1} (0), y_{1} (0)) = 0$
To satisfy the Additive Zero Axiom, $(x_{1}, y_{1}) + 0 = (x_{1}, y_{1})$ must be true. For this to be true, 0 would have to be (1,1)
Is this possible, or would we be able to say this is not a vector space?

timberwuf8r 2022-09-28

Instead of using arrows to represent a planar vector field, one sometimes uses families of curves called field lines. A curve $y = y (x)$ is a field line of the vector field F(x,y) if at each point $(x_{0}, y_{0})$ on the curve, $F (x_{0}, y_{0})$ is tangent to the curve.
Show that the field lines y=y(x) of a vector field $F (x, y) = P (x, y) i + Q (x, y) j$ are solutions to the differential equation $d y / d x = Q / P$
Find the field lines of $F (x, y) = y i + x j$

Karyme Gallardo2022-09-27

7. A projectile is fired with an initial speed of 65m / s at an angle of 35 degrees above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and () the velocity of the projectile 1.50 s after firing.

clovnerie0q 2022-09-27

I know that the equation of a hyperplane in n-dimensions is given by:
$w^{T} x + w_{0} = 0$
Where w is a vector that is perpendicular to the surface of the hyperplane and $w_{0}$ is a constant. I also know that if $w_{0} = 0$ , then the plane passes through the origin. My question is, what is the physical significance of the constant term $w_{0}$ . Is it the distance of the plane to the origin? If not then what is its physical significance?

Linda Peters 2022-09-27

Suppose $\vec{v}$ and $\vec{w}$ are two vectors parallel to the plane
$x + 2 y + 3 z = 7.$
Suppose furthermore that $\vec{v}$ is perpendicular to $\vec{w}$ ,
$‖ v ‖ = 3, ‖ w ‖ = 4.$
How would you go about answering this question? I always reach a dead end when I try to solve $\vec{v}$ and $\vec{w}$ that satisfy the given information, I just don't know how to go about it. I tried to draw the plane and two parallel vectors, then I know the length of the cross product would be 12. Then what do I do now? How could you solve this to find vectors $\vec{v}$ and $\vec{w}$ ?

tidones0r 2022-09-27

Say that we have n complex-valued vectors $z_{i}$ and we want to evaluate:
${‖ \sum_{i = 1}^{n} z_{i} ‖}_{2}^{2}$
Now, I know that for real vectors $x_{i}$ it holds:
${‖ \sum_{i = 1}^{n} x_{i} ‖}_{2}^{2} = \sum_{i = 1}^{n} {‖ x_{i} ‖}_{2}^{2} + \sum_{i \neq j} x_{i} \cdot x_{j}$
But when dealing with complex vectors, the last term with the inner product will be a complex number in general so I am not sure that this formula generalizes for this case. What am I missing?

2k1ablakrh0 2022-09-27

A,B,C,D are four points in the space and satisfy $∣ \vec{A B} ∣= 3, ∣ \vec{B C} ∣= 7, ∣ \vec{C D} ∣= 11$ and $∣ \vec{D A} ∣= 9$ . Then $\vec{A C} . \vec{B D}$ is
I know that $\vec{A B} + \vec{B C} + \vec{C A} + \vec{A D} = 0$ and also I observed that
$∣ \vec{A B} ∣^{2} + ∣ \vec{C D} ∣^{2} =∣ \vec{B C} ∣^{2} + ∣ \vec{D A} ∣^{2}$ . But couldn't get anything from it. Any hint?

Medenovgj 2022-09-27

How to find a position vector given several other position vectors with a given relationship?
Relative to the origin O , the position vector of A is $6 i + 6 j$ and the position vector of B is $12 i - 2 j$ . The point C lies on $\vec{A B}$ − such that $\vec{A C} = \frac{3}{4} \vec{A B}$ . Find the position vector of C.
As of now, I have found $\vec{A B}$ and $\vec{A C}$ :
$\vec{A B} = \vec{O B} - \vec{O A}$
$\vec{A B} = 12 i - 2 j - (6 i + 6 j)$
$\vec{A B} = 12 i - 2 j - 6 i - 6 j$
$∴ \vec{A B} = 6 i - 8 j$
$\vec{A C} = \frac{3}{4} \vec{A B}$
$\vec{A C} = \frac{3}{4} \cdot (6 i - 8 j)$
$\vec{A C} = \frac{3}{4} \cdot 2 (3 i - 4 j)$
$\vec{A C} = \frac{3}{2} \cdot (3 i - 4 j)$
$\vec{A C} = \frac{3 (3 i - 4 j)}{2}$
$∴ \vec{A C} = \frac{9 i - 12 j}{2}$
How would $\vec{O C}$ be found?

Camila Brandt 2022-09-27

Let $A$ be a matrix, and let 𝐱 and 𝐲 be linearly independent vectors such that 𝐀𝐱=𝐲,𝐀𝐲=𝐱+2𝐲.Then we have that $A^{5} x = a x + b y$ for some scalars 𝑎 and 𝑏. Find the ordered pair (a, b).
So far I have $A^{5} x = A^{4} y$
$A^{4} y = A^{3} x + 2 A^{3} y$

kjukks1234531 2022-09-26

Find the span of $v_{1}, v_{2}, . . ., v_{k}, w$ when $x_{1} \cdot v_{1} + x_{2} \cdot v_{2} + . . . + x_{k} \cdot v_{k} = w$ has no solution

Vectors in the Precalculus course are usually more challenging since there are different vectors examples that are always mentioned. For example, if you are majoring in Engineering disciplines, you will have to use more than one approach to explain the most efficient ways. Take a look at vectors practice problems that have been presented below. It will help you learn and find the answers that will let you see the best equation and graphs. At the same time, when you are dealing with vectors equations, do not forget about unknown coefficients as you are looking for combinations and various solutions.

Precalculus

Vector Examples, Equations, and Practice Problems