Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Let's analyze the given vector components and calculate their properties. We are given the vector components:
[tex]\[ \mathbf{v} = (r, s, t) = (5, -3, 8) \][/tex]
We'll calculate two main properties: the magnitude of the vector and the direction angle using the components [tex]\( r \)[/tex] and [tex]\( s \)[/tex].
### Step 1: Calculate the Magnitude
The magnitude [tex]\( \|\mathbf{v}\| \)[/tex] of a vector in 3-dimensional space [tex]\(\mathbf{v} = (r, s, t)\)[/tex] is given by the formula:
[tex]\[ \|\mathbf{v}\| = \sqrt{r^2 + s^2 + t^2} \][/tex]
Substituting the given values [tex]\( r = 5 \)[/tex], [tex]\( s = -3 \)[/tex], and [tex]\( t = 8 \)[/tex]:
[tex]\[ \|\mathbf{v}\| = \sqrt{5^2 + (-3)^2 + 8^2} = \sqrt{25 + 9 + 64} = \sqrt{98} \][/tex]
Evaluating [tex]\(\sqrt{98}\)[/tex]:
[tex]\[ \sqrt{98} \approx 9.899 \][/tex]
So, the magnitude of the vector is approximately:
[tex]\[ \|\mathbf{v}\| \approx 9.899 \][/tex]
### Step 2: Calculate the Direction Angle (θ)
The direction angle [tex]\( \theta \)[/tex] in the context of a 3D vector can often be understood as the angle between the projection of the vector on the [tex]\( xy \)[/tex]-plane and the positive [tex]\( x \)[/tex]-axis. This can be calculated as:
[tex]\[ \theta = \arctan\left(\frac{s}{r}\right) \][/tex]
Substituting [tex]\( r = 5 \)[/tex] and [tex]\( s = -3 \)[/tex] into the formula:
[tex]\[ \theta = \arctan\left(\frac{-3}{5}\right) \][/tex]
We convert this angle from radians to degrees:
[tex]\[ \theta \approx -30.964^\circ \][/tex]
So the direction angle is approximately:
[tex]\[ \theta \approx -30.964^\circ \][/tex]
### Consider the Given Options and Find the Closest Match
We compare our calculated magnitude and direction angle to the given options:
1. [tex]\(10.8, \theta = 56.3^\circ\)[/tex]
2. [tex]\(18.4, \theta = 119.4^\circ\)[/tex]
3. [tex]\(41.0, \theta = 77.3^\circ\)[/tex]
4. [tex]\(97.6, \theta = 24.2^\circ\)[/tex]
None of the options match our calculated values ([tex]\(\approx 9.899\)[/tex] for magnitude and [tex]\(-30.964^\circ\)[/tex] for direction). Thus, our result is:
[tex]\[ \rho \approx 9.899, \quad \theta \approx -30.964^\circ \][/tex]
[tex]\[ \mathbf{v} = (r, s, t) = (5, -3, 8) \][/tex]
We'll calculate two main properties: the magnitude of the vector and the direction angle using the components [tex]\( r \)[/tex] and [tex]\( s \)[/tex].
### Step 1: Calculate the Magnitude
The magnitude [tex]\( \|\mathbf{v}\| \)[/tex] of a vector in 3-dimensional space [tex]\(\mathbf{v} = (r, s, t)\)[/tex] is given by the formula:
[tex]\[ \|\mathbf{v}\| = \sqrt{r^2 + s^2 + t^2} \][/tex]
Substituting the given values [tex]\( r = 5 \)[/tex], [tex]\( s = -3 \)[/tex], and [tex]\( t = 8 \)[/tex]:
[tex]\[ \|\mathbf{v}\| = \sqrt{5^2 + (-3)^2 + 8^2} = \sqrt{25 + 9 + 64} = \sqrt{98} \][/tex]
Evaluating [tex]\(\sqrt{98}\)[/tex]:
[tex]\[ \sqrt{98} \approx 9.899 \][/tex]
So, the magnitude of the vector is approximately:
[tex]\[ \|\mathbf{v}\| \approx 9.899 \][/tex]
### Step 2: Calculate the Direction Angle (θ)
The direction angle [tex]\( \theta \)[/tex] in the context of a 3D vector can often be understood as the angle between the projection of the vector on the [tex]\( xy \)[/tex]-plane and the positive [tex]\( x \)[/tex]-axis. This can be calculated as:
[tex]\[ \theta = \arctan\left(\frac{s}{r}\right) \][/tex]
Substituting [tex]\( r = 5 \)[/tex] and [tex]\( s = -3 \)[/tex] into the formula:
[tex]\[ \theta = \arctan\left(\frac{-3}{5}\right) \][/tex]
We convert this angle from radians to degrees:
[tex]\[ \theta \approx -30.964^\circ \][/tex]
So the direction angle is approximately:
[tex]\[ \theta \approx -30.964^\circ \][/tex]
### Consider the Given Options and Find the Closest Match
We compare our calculated magnitude and direction angle to the given options:
1. [tex]\(10.8, \theta = 56.3^\circ\)[/tex]
2. [tex]\(18.4, \theta = 119.4^\circ\)[/tex]
3. [tex]\(41.0, \theta = 77.3^\circ\)[/tex]
4. [tex]\(97.6, \theta = 24.2^\circ\)[/tex]
None of the options match our calculated values ([tex]\(\approx 9.899\)[/tex] for magnitude and [tex]\(-30.964^\circ\)[/tex] for direction). Thus, our result is:
[tex]\[ \rho \approx 9.899, \quad \theta \approx -30.964^\circ \][/tex]
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
