Solved on Dec 09, 2023

Find a vector $\mathbf{v}$ with magnitude 4, where the $\mathbf{i}$ component is twice the $\mathbf{j}$ component.

STEP 1

Assumptions
1. The vector $\mathbf{v}$ has a magnitude of 4.
2. The component of $\mathbf{v}$ in the $\mathbf{i}$ direction is twice the component in the $\mathbf{j}$ direction.
3. The vector $\mathbf{v}$ can be represented as $\mathbf{v} = xi + yj$ , where $x$ and $y$ are the components in the $\mathbf{i}$ and $\mathbf{j}$ directions, respectively.

STEP 2

Let's denote the component in the $\mathbf{i}$ direction as $2y$ , since it is twice the component in the $\mathbf{j}$ direction, which we will denote as $y$ .
$\mathbf{v} = 2y\mathbf{i} + y\mathbf{j}$

STEP 3

The magnitude of a vector $\mathbf{v} = xi + yj$ is given by the square root of the sum of the squares of its components.
$|\mathbf{v}| = \sqrt{x^2 + y^2}$

STEP 4

Substitute $x = 2y$ into the magnitude formula, since the component in the $\mathbf{i}$ direction is twice the component in the $\mathbf{j}$ direction.
$|\mathbf{v}| = \sqrt{(2y)^2 + y^2}$

STEP 5

Simplify the expression under the square root by squaring the components.
$|\mathbf{v}| = \sqrt{4y^2 + y^2}$

STEP 6

Combine like terms under the square root.
$|\mathbf{v}| = \sqrt{5y^2}$

STEP 7

Take the square root of $5y^2$ .
$|\mathbf{v}| = \sqrt{5} |y|$

STEP 8

Since we know the magnitude of $\mathbf{v}$ is 4, we can set the above expression equal to 4.
$4 = \sqrt{5} |y|$

STEP 9

Solve for $|y|$ by dividing both sides of the equation by $\sqrt{5}$ .
$|y| = \frac{4}{\sqrt{5}}$

STEP 10

To simplify, multiply the numerator and denominator by $\sqrt{5}$ to rationalize the denominator.
$|y| = \frac{4\sqrt{5}}{5}$

STEP 11

Since $y$ can be either positive or negative, we consider both cases to find the possible vectors. Let's first consider the positive case.
$y = \frac{4\sqrt{5}}{5}$

STEP 12

Now, find the corresponding $x$ value using $x = 2y$ .
$x = 2 \left(\frac{4\sqrt{5}}{5}\right)$

STEP 13

Simplify the expression for $x$ .
$x = \frac{8\sqrt{5}}{5}$

STEP 14

Now we have the components for the case when $y$ is positive. The vector $\mathbf{v}$ is:
$\mathbf{v} = \frac{8\sqrt{5}}{5}\mathbf{i} + \frac{4\sqrt{5}}{5}\mathbf{j}$

STEP 15

Next, consider the case when $y$ is negative.
$y = -\frac{4\sqrt{5}}{5}$

STEP 16

Find the corresponding $x$ value using $x = 2y$ .
$x = 2 \left(-\frac{4\sqrt{5}}{5}\right)$

STEP 17

Simplify the expression for $x$ .
$x = -\frac{8\sqrt{5}}{5}$

STEP 18

Now we have the components for the case when $y$ is negative. The vector $\mathbf{v}$ is:
$\mathbf{v} = -\frac{8\sqrt{5}}{5}\mathbf{i} - \frac{4\sqrt{5}}{5}\mathbf{j}$

STEP 19

We have found two vectors that satisfy the given conditions, one with positive components and one with negative components.
The vectors are:
$\mathbf{v} = \frac{8\sqrt{5}}{5}\mathbf{i} + \frac{4\sqrt{5}}{5}\mathbf{j}$
and
$\mathbf{v} = -\frac{8\sqrt{5}}{5}\mathbf{i} - \frac{4\sqrt{5}}{5}\mathbf{j}$
These vectors have a magnitude of 4 and their component in the $\mathbf{i}$ direction is twice the component in the $\mathbf{j}$ direction.

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Find a vector $\mathbf{v}$ with magnitude 4, where the $\mathbf{i}$ component is twice the $\mathbf{j}$ component.

STEP 1

STEP 2

Let's denote the component in the $\mathbf{i}$ direction as $2y$ , since it is twice the component in the $\mathbf{j}$ direction, which we will denote as $y$ .
$\mathbf{v} = 2y\mathbf{i} + y\mathbf{j}$

STEP 3

The magnitude of a vector $\mathbf{v} = xi + yj$ is given by the square root of the sum of the squares of its components.
$|\mathbf{v}| = \sqrt{x^2 + y^2}$

STEP 4

Substitute $x = 2y$ into the magnitude formula, since the component in the $\mathbf{i}$ direction is twice the component in the $\mathbf{j}$ direction.
$|\mathbf{v}| = \sqrt{(2y)^2 + y^2}$

STEP 5

Simplify the expression under the square root by squaring the components.
$|\mathbf{v}| = \sqrt{4y^2 + y^2}$

STEP 6

Combine like terms under the square root.
$|\mathbf{v}| = \sqrt{5y^2}$

STEP 7

Take the square root of $5y^2$ .
$|\mathbf{v}| = \sqrt{5} |y|$

STEP 8

Since we know the magnitude of $\mathbf{v}$ is 4, we can set the above expression equal to 4.
$4 = \sqrt{5} |y|$

STEP 9

Solve for $|y|$ by dividing both sides of the equation by $\sqrt{5}$ .
$|y| = \frac{4}{\sqrt{5}}$

STEP 10

To simplify, multiply the numerator and denominator by $\sqrt{5}$ to rationalize the denominator.
$|y| = \frac{4\sqrt{5}}{5}$

STEP 11

Since $y$ can be either positive or negative, we consider both cases to find the possible vectors. Let's first consider the positive case.
$y = \frac{4\sqrt{5}}{5}$

STEP 12

Now, find the corresponding $x$ value using $x = 2y$ .
$x = 2 \left(\frac{4\sqrt{5}}{5}\right)$

STEP 13

Simplify the expression for $x$ .
$x = \frac{8\sqrt{5}}{5}$

STEP 14

Now we have the components for the case when $y$ is positive. The vector $\mathbf{v}$ is:
$\mathbf{v} = \frac{8\sqrt{5}}{5}\mathbf{i} + \frac{4\sqrt{5}}{5}\mathbf{j}$

STEP 15

Next, consider the case when $y$ is negative.
$y = -\frac{4\sqrt{5}}{5}$

STEP 16

Find the corresponding $x$ value using $x = 2y$ .
$x = 2 \left(-\frac{4\sqrt{5}}{5}\right)$

STEP 17

Simplify the expression for $x$ .
$x = -\frac{8\sqrt{5}}{5}$

STEP 18

Now we have the components for the case when $y$ is negative. The vector $\mathbf{v}$ is:
$\mathbf{v} = -\frac{8\sqrt{5}}{5}\mathbf{i} - \frac{4\sqrt{5}}{5}\mathbf{j}$

STEP 19

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Find a vector v\mathbf{v}v with magnitude 4, where the i\mathbf{i}i component is twice the j\mathbf{j}j component.

STEP 1

STEP 2

Let's denote the component in the i\mathbf{i}i direction as 2y2y2y, since it is twice the component in the j\mathbf{j}j direction, which we will denote as yyy.v=2yi+yj\mathbf{v} = 2y\mathbf{i} + y\mathbf{j}v=2yi+yj

STEP 3

The magnitude of a vector v=xi+yj\mathbf{v} = xi + yjv=xi+yj is given by the square root of the sum of the squares of its components.∣v∣=x2+y2|\mathbf{v}| = \sqrt{x^2 + y^2}∣v∣=x2+y2​

STEP 4

Substitute x=2yx = 2yx=2y into the magnitude formula, since the component in the i\mathbf{i}i direction is twice the component in the j\mathbf{j}j direction.∣v∣=(2y)2+y2|\mathbf{v}| = \sqrt{(2y)^2 + y^2}∣v∣=(2y)2+y2​

STEP 5

Simplify the expression under the square root by squaring the components.∣v∣=4y2+y2|\mathbf{v}| = \sqrt{4y^2 + y^2}∣v∣=4y2+y2​

STEP 6

Combine like terms under the square root.∣v∣=5y2|\mathbf{v}| = \sqrt{5y^2}∣v∣=5y2​

STEP 7

Take the square root of 5y25y^25y2.∣v∣=5∣y∣|\mathbf{v}| = \sqrt{5} |y|∣v∣=5​∣y∣

STEP 8

Since we know the magnitude of v\mathbf{v}v is 4, we can set the above expression equal to 4.4=5∣y∣4 = \sqrt{5} |y|4=5​∣y∣

STEP 9

Solve for ∣y∣|y|∣y∣ by dividing both sides of the equation by 5\sqrt{5}5​.∣y∣=45|y| = \frac{4}{\sqrt{5}}∣y∣=5​4​

STEP 10

To simplify, multiply the numerator and denominator by 5\sqrt{5}5​ to rationalize the denominator.∣y∣=455|y| = \frac{4\sqrt{5}}{5}∣y∣=545​​

STEP 11

Since yyy can be either positive or negative, we consider both cases to find the possible vectors. Let's first consider the positive case.y=455y = \frac{4\sqrt{5}}{5}y=545​​

STEP 12

Now, find the corresponding xxx value using x=2yx = 2yx=2y.x=2(455)x = 2 \left(\frac{4\sqrt{5}}{5}\right)x=2(545​​)

STEP 13

Simplify the expression for xxx.x=855x = \frac{8\sqrt{5}}{5}x=585​​

STEP 14

Now we have the components for the case when yyy is positive. The vector v\mathbf{v}v is:v=855i+455j\mathbf{v} = \frac{8\sqrt{5}}{5}\mathbf{i} + \frac{4\sqrt{5}}{5}\mathbf{j}v=585​​i+545​​j

STEP 15

Next, consider the case when yyy is negative.y=−455y = -\frac{4\sqrt{5}}{5}y=−545​​

STEP 16

Find the corresponding xxx value using x=2yx = 2yx=2y.x=2(−455)x = 2 \left(-\frac{4\sqrt{5}}{5}\right)x=2(−545​​)

STEP 17

Simplify the expression for xxx.x=−855x = -\frac{8\sqrt{5}}{5}x=−585​​

STEP 18

Now we have the components for the case when yyy is negative. The vector v\mathbf{v}v is:v=−855i−455j\mathbf{v} = -\frac{8\sqrt{5}}{5}\mathbf{i} - \frac{4\sqrt{5}}{5}\mathbf{j}v=−585​​i−545​​j

STEP 19

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Find a vector $\mathbf{v}$ with magnitude 4, where the $\mathbf{i}$ component is twice the $\mathbf{j}$ component.

Let's denote the component in the $\mathbf{i}$ direction as $2y$ , since it is twice the component in the $\mathbf{j}$ direction, which we will denote as $y$ .
$\mathbf{v} = 2y\mathbf{i} + y\mathbf{j}$

The magnitude of a vector $\mathbf{v} = xi + yj$ is given by the square root of the sum of the squares of its components.
$|\mathbf{v}| = \sqrt{x^2 + y^2}$

Substitute $x = 2y$ into the magnitude formula, since the component in the $\mathbf{i}$ direction is twice the component in the $\mathbf{j}$ direction.
$|\mathbf{v}| = \sqrt{(2y)^2 + y^2}$

Simplify the expression under the square root by squaring the components.
$|\mathbf{v}| = \sqrt{4y^2 + y^2}$

Combine like terms under the square root.
$|\mathbf{v}| = \sqrt{5y^2}$

Take the square root of $5y^2$ .
$|\mathbf{v}| = \sqrt{5} |y|$

Since we know the magnitude of $\mathbf{v}$ is 4, we can set the above expression equal to 4.
$4 = \sqrt{5} |y|$

Solve for $|y|$ by dividing both sides of the equation by $\sqrt{5}$ .
$|y| = \frac{4}{\sqrt{5}}$

To simplify, multiply the numerator and denominator by $\sqrt{5}$ to rationalize the denominator.
$|y| = \frac{4\sqrt{5}}{5}$

Since $y$ can be either positive or negative, we consider both cases to find the possible vectors. Let's first consider the positive case.
$y = \frac{4\sqrt{5}}{5}$

Now, find the corresponding $x$ value using $x = 2y$ .
$x = 2 \left(\frac{4\sqrt{5}}{5}\right)$

Simplify the expression for $x$ .
$x = \frac{8\sqrt{5}}{5}$

Now we have the components for the case when $y$ is positive. The vector $\mathbf{v}$ is:
$\mathbf{v} = \frac{8\sqrt{5}}{5}\mathbf{i} + \frac{4\sqrt{5}}{5}\mathbf{j}$

Next, consider the case when $y$ is negative.
$y = -\frac{4\sqrt{5}}{5}$

Find the corresponding $x$ value using $x = 2y$ .
$x = 2 \left(-\frac{4\sqrt{5}}{5}\right)$

Simplify the expression for $x$ .
$x = -\frac{8\sqrt{5}}{5}$

Now we have the components for the case when $y$ is negative. The vector $\mathbf{v}$ is:
$\mathbf{v} = -\frac{8\sqrt{5}}{5}\mathbf{i} - \frac{4\sqrt{5}}{5}\mathbf{j}$