Sure, let's approach this problem step-by-step.
1. Vectors and angle:
- Vector [tex]\( A = 2\mathbf{i} - \mathbf{j} + \mathbf{k} \)[/tex]
- Vector [tex]\( B = \mathbf{i} + 3\mathbf{j} - 4\mathbf{k} \)[/tex]
- Angle between [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is [tex]\( 60^\circ \)[/tex]
2. Find magnitudes of vectors [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
The magnitude of a vector [tex]\( \mathbf{V} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \)[/tex] is given by
[tex]\[
|\mathbf{V}| = \sqrt{a^2 + b^2 + c^2}
\][/tex]
- For vector [tex]\( A \)[/tex]:
[tex]\[
|A| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \approx 2.449
\][/tex]
- For vector [tex]\( B \)[/tex]:
[tex]\[
|B| = \sqrt{1^2 + 3^2 + (-4)^2} = \sqrt{1 + 9 + 16} = \sqrt{26} \approx 5.099
\][/tex]
3. Calculate the dot product of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
The dot product [tex]\( A \cdot B \)[/tex] is given by:
[tex]\[
A \cdot B = (2)(1) + (-1)(3) + (1)(-4) = 2 - 3 - 4 = -5
\][/tex]
4. Determine the cosine of the angle:
Given the angle [tex]\( \theta = 60^\circ \)[/tex], we can find [tex]\( \cos(60^\circ) \)[/tex]:
[tex]\[
\cos(60^\circ) = \frac{1}{2}
\][/tex]
5. Use the resultant vector magnitude formula:
The magnitude of the resultant vector [tex]\( R \)[/tex] formed by vectors [tex]\( A \)[/tex] and [tex]\( B \)[/tex] at an angle [tex]\( \theta \)[/tex] is given by:
[tex]\[
|R| = \sqrt{|A|^2 + |B|^2 + 2|A||B|\cos(\theta)}
\][/tex]
Plug in the values:
[tex]\[
|R| = \sqrt{(2.449)^2 + (5.099)^2 + 2(2.449)(5.099)\left(\frac{1}{2}\right)}
\][/tex]
Calculating terms step-by-step:
[tex]\[
(2.449)^2 \approx 6
\][/tex]
[tex]\[
(5.099)^2 \approx 26
\][/tex]
[tex]\[
2(2.449)(5.099)\left(\frac{1}{2}\right) \approx 12.5
\][/tex]
Therefore:
[tex]\[
|R| = \sqrt{6 + 26 + 12.5} = \sqrt{44.5} \approx 6.670
\][/tex]
So, the magnitude of the resultant vector is approximately [tex]\( 6.670 \)[/tex].
