Example: Show that vector [tex]\(\vec{A} = 3i + 6j - 2k\)[/tex] and [tex]\(\vec{B} = 4i - j + 3k\)[/tex] are mutually perpendicular.

Solution:

[tex]\[
\vec{A} \cdot \vec{B} = \vec{A} \cdot \vec{B} \cos \theta
\][/tex]
[tex]\[
\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} = \frac{(3i + 6j - 2k) \cdot (4i - j + 3k)}{|\vec{A}||\vec{B}|}
\][/tex]

Calculate the dot product:

[tex]\[
\vec{A} \cdot \vec{B} = (3 \times 4) + (6 \times -1) + (-2 \times 3)
\][/tex]
[tex]\[
= 12 - 6 - 6
\][/tex]
[tex]\[
= 0
\][/tex]

Since [tex]\(\vec{A} \cdot \vec{B} = 0\)[/tex],

[tex]\[
\cos \theta = 0 \Rightarrow \theta = 90^\circ
\][/tex]

Therefore, the vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] are mutually perpendicular.

---

Assignment:
A particle moves from a point [tex]\((35, -4, -2) \, \text{m}\)[/tex] to another point [tex]\((5, -6, 2) \, \text{m}\)[/tex] under the influence of a force [tex]\(\vec{F} = (-3i + 4j + 4k) \, \text{N}\)[/tex]. Calculate the work done by the force.



Answer :

Sure, let's solve the given problem step by step:

### Problem:
A particle moves from a point [tex]\((35, -4, -2)\)[/tex] to another point [tex]\((5, -6, 2)\)[/tex] under the influence of a force [tex]\(\vec{F} = -3\hat{i} + 4\hat{j} + 4\hat{k} \, \text{N}\)[/tex]. Calculate the work done by the force.

### Solution:

1. Identify Initial and Final Points:
We have the initial point [tex]\((35, -4, -2)\)[/tex] and the final point [tex]\((5, -6, 2)\)[/tex].

2. Determine the Displacement Vector:
The displacement vector [tex]\(\vec{d}\)[/tex] can be found by subtracting the coordinates of the initial point from the coordinates of the final point:

[tex]\[ \vec{d} = \text{final point} - \text{initial point} \][/tex]

Given:

[tex]\[ \text{initial point} = (35, -4, -2) \][/tex]

[tex]\[ \text{final point} = (5, -6, 2) \][/tex]

Therefore:

[tex]\[ \vec{d} = (5 - 35, -6 + 4, 2 + 2) \][/tex]

Simplifying:

[tex]\[ \vec{d} = (-30, -2, 4) \][/tex]

Thus, the displacement vector [tex]\(\vec{d} = -30\hat{i} - 2\hat{j} + 4\hat{k}\)[/tex].

3. Force Vector:
Given the force vector:

[tex]\[ \vec{F} = -3\hat{i} + 4\hat{j} + 4\hat{k} \, \text{N} \][/tex]

4. Calculate the Work Done:
Work done by a force [tex]\(\vec{F}\)[/tex] over a displacement [tex]\(\vec{d}\)[/tex] is given by the dot product of [tex]\(\vec{F}\)[/tex] and [tex]\(\vec{d}\)[/tex]:

[tex]\[ W = \vec{F} \cdot \vec{d} \][/tex]

Substituting the given vectors:

[tex]\[ \vec{F} = (-3, 4, 4) \][/tex]

[tex]\[ \vec{d} = (-30, -2, 4) \][/tex]

The dot product [tex]\(\vec{F} \cdot \vec{d}\)[/tex] is:

[tex]\[ W = (-3 \times -30) + (4 \times -2) + (4 \times 4) \][/tex]

Simplifying each term:

[tex]\[ W = 90 + (-8) + 16 \][/tex]

Summing these:

[tex]\[ W = 90 - 8 + 16 \][/tex]

[tex]\[ W = 98 \, \text{J} \][/tex]

Therefore, the work done by the force is [tex]\(98 \, \text{J}\)[/tex].