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(6) The magnitu 0A OE 5 J
Sa RE tie eiiyectos 04,08 and OC in the given figure are equal. The direction of 04 + OB — OC with
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Question:
step by step NREL oe Sep SS SOAR Lp \C€) must not be zero (d) All of above are possible
(6) The magnitu 0A OE 5 J
Sa RE tie eiiyectos 04,08 and OC in the given figure are equal. The direction of 04 + OB — OC with
y ul OTIS
Cc i AS ig Se - - PED
i: ~ \N\N.-
45 A30° A — a
oR ey OR — a
i ga Ne— ‘4
i 8 rine mp rn us
ch) Gon SN Bey "NS oR TEE
_3 A—V3—VZ] 1 (3-14V2) -1 (3-142) va:
@ wes OEE s © wn avrg @ tan fete - [36% of
This is AI generated Response.
Asked by: me
Created at: 2025-05-30 08:32:14
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Let's solve the problem step by step:
### **Given:**
- Magnitude of \( \vec{OA}, \vec{OB}, \vec{OC} \) are equal.
- Find the direction of \( \vec{OA} + \vec{OB} - \vec{OC} \) with the x-axis.
Let the magnitude of each vector = \( r \) (let's take \( r = 1 \) for simplicity).
---
### **Step 1: Write the Vectors in Component Form**
From the diagram:
- \( \vec{OA} \) makes a 30° angle with x-axis (in first quadrant).
- \( \vec{OB} \) makes a 60° angle with negative x-axis (in fourth quadrant below x-axis).
- \( \vec{OC} \) makes a 45° angle with positive y-axis (second quadrant).
Let’s break them into components:
#### **\( \vec{OA} \)**
Angle with x-axis = 30°
\[
\vec{OA} = r(\cos 30^\circ \, \hat{i} + \sin 30^\circ \, \hat{j})
\]
\[
= (\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j})
\]
#### **\( \vec{OB} \)**
Angle with x-axis = 60°, but it's in the fourth quadrant (so, x positive, y negative).
\[
\vec{OB} = r(\cos 60^\circ \, \hat{i} - \sin 60^\circ \, \hat{j})
\]
\[
= (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})
\]
#### **\( \vec{OC} \)**
Angle with x-axis = 45°, but it's in the second quadrant (so, x negative, y positive).
\[
\vec{OC} = r(-\cos 45^\circ \, \hat{i} + \sin 45^\circ \, \hat{j})
\]
\[
= (-\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})
\]
---
### **Step 2: Compute \( \vec{OA} + \vec{OB} - \vec{OC} \)**
Sum the i and j components:
#### **i-component:**
\[
\left(\frac{\sqrt{3}}{2} + \frac{1}{2} - \left(-\frac{1}{\sqrt{2}}\right)\right)
= \frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}}
\]
#### **j-component:**
\[
\left(\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\right)
= \frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}
\]
---
### **Step 3: Direction with x-axis**
Let \( \theta \) be the angle with the x-axis:
\[
\tan\theta = \frac{\text{y-component}}{\text{x-component}}
= \frac{\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}}}
\]
Let's simplify numerator and denominator:
#### **Numerator:**
\[
\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}
\]
#### **Denominator:**
\[
\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}}
\]
Multiply numerator and denominator by 2 to clear denominators:
Numerator:
\[
1 - \sqrt{3} - \sqrt{2}
\]
Denominator:
\[
\sqrt{3} + 1 + \sqrt{2}
\]
Now, match with the options:
- Option (a): \(\tan^{-1}\left(\frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}\right)\)
**This matches exactly!**
---
## **Final Answer**
\[
\boxed{
\text{Option (a): } \tan^{-1}\left(\frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}\right)
}
\]
Let me know if you need any further explanation!
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