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Find the following vectors. The vector that is 3 times \langle 3,-5\rangle plus -9 times \langle 6,0\rangle

Short Answer

Expert verified
Answer: The resulting vector is <-45, -15>.

Step by step solution

01

Multiply the first vector by 3

To multiply the vector \langle 3, -5\rangle by 3, we multiply each of its components by 3: \[3\langle 3, -5\rangle = \langle 3(3), 3(-5) \rangle = \langle 9, -15 \rangle\]
02

Multiply the second vector by -9

To multiply the vector \langle 6, 0\rangle by -9, we multiply each of its components by -9: \[-9\langle 6, 0\rangle = \langle -9(6), -9(0) \rangle = \langle -54, 0 \rangle\]
03

Add the resulting vectors from steps 1 and 2

Now add the resulting vectors from steps 1 and 2 to obtain the final vector: \[\langle 9, -15\rangle + \langle -54, 0 \rangle = \langle 9 + (-54), -15 + 0 \rangle = \langle -45, -15 \rangle\] The vector that is 3 times \langle 3, -5\rangle plus -9 times \langle 6, 0\rangle is \(\boxed{\langle -45, -15\rangle}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Multiplication
Vector multiplication involves multiplying each component of a vector by a scalar (a number). This process scales the vector, either increasing or decreasing its magnitude. When we multiply the vector \( \langle 3, -5\rangle \) by 3, each component is multiplied individually:
  • Multiply the x-component: \( 3 \times 3 = 9 \)
  • Multiply the y-component: \( 3 \times (-5) = -15 \)
This gives us the new vector \( \langle 9, -15\rangle \). Similarly, for the vector \( \langle 6, 0\rangle \) multiplied by \(-9\), we get:
  • \(-9 \times 6 = -54\)
  • \(-9 \times 0 = 0\)
Thus, the vector becomes \( \langle -54, 0\rangle \). Multiplying vectors by scalars alters their size but not their direction, unless the scalar is negative. Then, the vector points in the opposite direction.
Vector Addition
Adding vectors involves adding the corresponding components of the vectors together. This operation results in a new vector. In our initial exercise, after finding:
  • \( \langle 9, -15\rangle \)
  • \( \langle -54, 0\rangle \)
We add these together by summing their x-components and y-components separately:
  • X-component: \( 9 + (-54) = -45 \)
  • Y-component: \( -15 + 0 = -15 \)
Thus, the resulting vector is \( \langle -45, -15\rangle \). This combines the effects of both vectors, taking into account their directions and magnitudes, resulting in a vector that represents the cumulative effect of both.
Mathematical Problem Solving
Approaching mathematical problems involving vectors requires a clear understanding of vector operations like multiplication and addition. To solve such problems effectively:
  • Start by identifying the vectors and operations required (like scalar multiplication).
  • Perform operations step by step, focusing on one vector at a time.
  • Be careful with signs, especially when multiplying by negative scalars, as these can change the vector's direction.
  • After computing each step, check your calculations to ensure they are correct.
  • Finally, combine your results, like through vector addition, to find your answer.
This process helps break down complicated problems into manageable steps, making it easier to understand and solve vector-related problems.

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Most popular questions from this chapter

Suppose water flows in a thin sheet over the \(x y\) -plane with a uniform velocity given by the vector \(\mathbf{v}=\langle 1,2\rangle ;\) this means that at all points of the plane, the velocity of the water has components \(1 \mathrm{m} / \mathrm{s}\) in the \(x\) -direction and \(2 \mathrm{m} / \mathrm{s}\) in the \(y\) -direction (see figure). Let \(C\) be an imaginary unit circle (that does not interfere with the flow). a. Show that at the point \((x, y)\) on the circle \(C\) the outwardpointing unit vector normal to \(C\) is \(\mathbf{n}=\langle x, y\rangle\) b. Show that at the point \((\cos \theta, \sin \theta)\) on the circle \(C\) the outward-pointing unit vector normal to \(C\) is also $$ \mathbf{n}=\langle\cos \theta, \sin \theta\rangle $$ c. Find all points on \(C\) at which the velocity is normal to \(C\). d. Find all points on \(C\) at which the velocity is tangential to \(C\). e. At each point on \(C\) find the component of \(v\) normal to \(C\) Express the answer as a function of \((x, y)\) and as a function of \(\theta\) f. What is the net flow through the circle? That is, does water accumulate inside the circle?

The points \(P, Q, R,\) and \(S,\) joined by the vectors \(\mathbf{u}, \mathbf{v}, \mathbf{w},\) and \(\mathbf{x},\) are the vertices of a quadrilateral in \(\mathrm{R}^{3}\). The four points needn't lie in \(a\) plane (see figure). Use the following steps to prove that the line segments joining the midpoints of the sides of the quadrilateral form a parallelogram. The proof does not use a coordinate system. a. Use vector addition to show that \(\mathbf{u}+\mathbf{v}=\mathbf{w}+\mathbf{x}\) b. Let \(m\) be the vector that joins the midpoints of \(P Q\) and \(Q R\) Show that \(\mathbf{m}=(\mathbf{u}+\mathbf{v}) / 2\) c. Let n be the vector that joins the midpoints of \(P S\) and \(S R\). Show that \(\mathbf{n}=(\mathbf{x}+\mathbf{w}) / 2\) d. Combine parts (a), (b), and (c) to conclude that \(\mathbf{m}=\mathbf{n}\) e. Explain why part (d) implies that the line segments joining the midpoints of the sides of the quadrilateral form a parallelogram.

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Write the vector \langle 2,-6\rangle in terms of \(\mathbf{I}\) and \(\mathbf{J}\).

A race Two people travel from \(P(4,0)\) to \(Q(-4,0)\) along the paths given by $$ \begin{aligned} \mathbf{r}(t) &=(4 \cos (\pi t / 8), 4 \sin (\pi t / 8)\rangle \text { and } \\\ \mathbf{R}(t) &=\left(4-t,(4-t)^{2}-16\right) \end{aligned} $$ a. Graph both paths between \(P\) and \(Q\) b. Graph the speeds of both people between \(P\) and \(Q\) c. Who arrives at \(Q\) first?

Torsion formula Show that the formula defining the torsion, \(\tau=-\frac{d \mathbf{B}}{d s} \cdot \mathbf{N},\) is equivalent to \(\tau=-\frac{1}{|\mathbf{v}|} \frac{d \mathbf{B}}{d t} \cdot \mathbf{N} .\) The second formula is generally easier to use.

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