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Chapter 12: Problem 34

Find a vector of magnitude 3 in the direction opposite to the direction of \(\mathbf{v}=(1 / 2) \mathbf{i}-(1 / 2) \mathbf{j}-(1 / 2) \mathbf{k}\).

Short Answer

Expert verified
The vector is \( \mathbf{w} = -\sqrt{3} \mathbf{i} + \sqrt{3} \mathbf{j} + \sqrt{3} \mathbf{k} \).

Step by step solution

01

Calculate the Magnitude of Vector v

The first step is to calculate the magnitude of the given vector \( \mathbf{v} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \).The magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is calculated using the formula: \[ |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \]For vector \( \mathbf{v} \):\[ |\mathbf{v}| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} \]\[ = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{4}} \]\[ = \sqrt{\frac{3}{4}} \]\[ = \frac{\sqrt{3}}{2} \]
02

Determine the Unit Vector in Opposite Direction

To find the vector in the opposite direction of \( \mathbf{v} \), first find the unit vector in the opposite direction.Unit vector \( \mathbf{u} \) in the direction opposite \( \mathbf{v} \) is given by:\[ \mathbf{u} = -\frac{\mathbf{v}}{|\mathbf{v}|} \]So the unit vector opposite to \( \mathbf{v} \) is:\[ \mathbf{u} = -\frac{1}{\frac{\sqrt{3}}{2}} \left( \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \right) \]\[ = -\frac{2}{\sqrt{3}} \left( \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \right) \]\[ = -\frac{1}{\sqrt{3}}(\mathbf{i} - \mathbf{j} - \mathbf{k}) \]\[ = -\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \]
03

Scale the Unit Vector to the Desired Magnitude

Now, we want to scale the unit vector \( \mathbf{u} \) to have a magnitude of 3.The scaled vector \( \mathbf{w} \) is given by multiplying the unit vector by the desired magnitude:\[ \mathbf{w} = 3 \mathbf{u} \]Substitute \( \mathbf{u} \):\[ \mathbf{w} = 3 \left( -\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \right) \]\[ = -\frac{3}{\sqrt{3}} \mathbf{i} + \frac{3}{\sqrt{3}} \mathbf{j} + \frac{3}{\sqrt{3}} \mathbf{k} \]Simplifying, since \( \frac{3}{\sqrt{3}} = \sqrt{3} \):\[ \mathbf{w} = -\sqrt{3} \mathbf{i} + \sqrt{3} \mathbf{j} + \sqrt{3} \mathbf{k} \]
04

Write the Final Vector

This is the vector of magnitude 3 in the direction opposite to the vector \( \mathbf{v} \):\[ \mathbf{w} = -\sqrt{3} \mathbf{i} + \sqrt{3} \mathbf{j} + \sqrt{3} \mathbf{k} \]

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Key Concepts

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

Unit Vector
A unit vector is a vector that has a magnitude of exactly 1. It is often used to indicate direction without regard to magnitude. To find a unit vector in the same direction as a given vector, you simply divide each component of the vector by its magnitude. For instance, if we have a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the unit vector \( \mathbf{u} \) is given by:\[ \mathbf{u} = \frac{1}{|\mathbf{v}|} \mathbf{v} = \frac{1}{|\mathbf{v}|} (a \mathbf{i} + b \mathbf{j} + c \mathbf{k}) \]
  • The unit vector retains the direction of the original vector.
  • It is especially useful for finding vectors with different magnitudes in the same direction.
Remember, the magnitude of a unit vector is always 1. Converting a vector to a unit vector does not change its direction, only the length is adjusted.
Opposite Direction
Sometimes, you may need a vector that points in the opposite direction of a given vector. To achieve this, you negate the vector's components, which flips its direction. For example, if \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the vector in the opposite direction, \(-\mathbf{v}\), is:\[ -\mathbf{v} = -a \mathbf{i} - b \mathbf{j} - c \mathbf{k} \]
  • Negating a vector changes its direction by 180 degrees.
  • Magnitude remains the same; only direction is reversed.
When combined with making a unit vector, you can find a unit vector in the opposite direction, useful for many physics and engineering problems.
Vector Scaling
Vector scaling involves adjusting the magnitude of a vector while keeping its direction unchanged. This is achieved by multiplying the vector by a scalar (a constant number). The resulting vector is larger or smaller, but still points in the same direction, or the opposite direction if the scalar is negative. To scale a vector \( \mathbf{u} \) by a scalar \( k \), the formula is:\[ \mathbf{w} = k \mathbf{u} \]
  • The direction of the vector remains constant if \( k > 0 \).
  • If \( k \) is negative, the direction gets flipped.
  • Magnitude of the new vector is the absolute value of \( k \) multiplied by the magnitude of \( \mathbf{u} \).
In problems, after finding the unit vector, you can scale it to suit your needs, allowing for precise control over both direction and magnitude.

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Most popular questions from this chapter

Sketch the surfaces. $$16 y^{2}+9 z^{2}=4 x^{2}$$

Express each vector as a product of its length and direction. $$5 \mathbf{k}$$

Find a. the direction of \(\overrightarrow{P_{1} P_{2}}\) and b. the midpoint of line segment \(P_{1} P_{2}\) $$P_{1}(-1,1,5) \quad P_{2}(2,5,0)$$

a. Find the volume of the solid bounded by the hyperboloid $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1 $$ and the planes \(z=0\) and \(z=h, h>0\). b. Express your answer in part (a) in terms of \(h\) and the areas \(A_{0}\) and \(A_{h}\) of the regions cut by the hyperboloid from the planes \(z=0\) and \(z=h\) c. Show that the volume in part (a) is also given by the formula $$ V=\frac{h}{6}\left(A_{0}+4 A_{m}+A_{h}\right), $$ where \(A_{m}\) is the area of the region cut by the hyperboloid from the plane \(z=h / 2\).

Find the point on the sphere \(x^{2}+(y-3)^{2}+(z+5)^{2}=4\) nearest a. the \(x y\) -plane. b. the point (0,7,-5)
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