The magnitude of vectors OA, OB, and OC in the given figure are equal. The direction of OA + OB - OC

Q: Why do we resolve vectors into components instead of using geometry?

When dealing with more than two vectors, geometric methods like the triangle law become very complex. Resolving them into orthogonal (perpendicular) components turns vector addition into simple algebraic addition.

Q: Why is the $x$-component of $\vec{OC}$ negative?

Vector $\vec{OC}$ lies in the second quadrant. In the second quadrant, the $x$-coordinates are negative while the $y$-coordinates are positive. Hence, its $x$-component is $-m\cos 45^\circ$.

Q: What does subtracting $\vec{OC}$ actually mean?

Subtracting a vector is identical to adding a vector that has the exact same magnitude but points in the completely opposite direction. So, $-\vec{OC}$ points into the fourth quadrant.

Q: How did we get $\sqrt{2}$ in the numerator when the component had $1/\sqrt{2}$?

To add fractions like $\frac{1}{2}$ and $\frac{1}{\sqrt{2}}$, we took a common denominator of $2$. Since $\frac{1}{\sqrt{2}}$ is mathematically equivalent to $\frac{\sqrt{2}}{2}$, it seamlessly combines with the other terms.

Q: Does the magnitude $m$ affect the final direction?

No! Since all vectors have the same magnitude $m$, it factors out completely when we take the ratio of the $y$-component to the $x$-component to find $\tan \theta$.

Question

The magnitude of vectors $\vec{OA}$, $\vec{OB}$, and $\vec{OC}$ in the given figure are equal. The direction of $\vec{OA} + \vec{OB} - \vec{OC}$ with X-axis will be

Options:

(A) $	an^{-1}\left(\frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}ight)$

(B) $	an^{-1}\left(\frac{\sqrt{3}-1+\sqrt{2}}{1+\sqrt{3}-\sqrt{2}}ight)$

(C) $	an^{-1}\left(\frac{\sqrt{3}-1+\sqrt{2}}{1-\sqrt{3}+\sqrt{2}}ight)$

(D) $	an^{-1}\left(\frac{1+\sqrt{3}-\sqrt{2}}{1-\sqrt{3}-\sqrt{2}}ight)$

Sigma Percentile · Accepted Answer

### The Visual Setup
Imagine standing at the origin of a coordinate system, holding three ropes pulling in different directions. This is exactly what our vectors $\vec{OA}$, $\vec{OB}$, and $\vec{OC}$ represent. They all have the same magnitude, let's call it $m$, but they point at different angles.

Vector $\vec{OA}$ points into the first quadrant at $30^\circ$. Vector $\vec{OB}$ dips into the fourth quadrant at $60^\circ$ below the X-axis. Finally, vector $\vec{OC}$ shoots into the second quadrant, making a $45^\circ$ angle with the negative X-axis.

### The Power of Resolution
Instead of wrestling with complex geometry like the triangle or parallelogram law, we use the ultimate strategy: **Vector Resolution**. By breaking each vector into its horizontal ($x$) and vertical ($y$) components, we turn a tricky geometric problem into simple algebra.

Any vector can be expressed as:
$$\vec{V} = V \cos	heta \hat{i} + V \sin	heta \hat{j}$$

### Breaking Down the Vectors
Let's analyze each vector one by one:

For $\vec{OA}$, the angle is $30^\circ$:
$$\vec{OA} = m\left(\cos 30^\circ \hat{i} + \sin 30^\circ \hat{j}ight) = m\left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}ight)$$

For $\vec{OB}$, the angle is $-60^\circ$ (since it's below the X-axis):
$$\vec{OB} = m\left(\cos(-60^\circ) \hat{i} + \sin(-60^\circ) \hat{j}ight) = m\left(\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}ight)$$

For $\vec{OC}$, it lies in the second quadrant. Its $x$-component is negative, and its $y$-component is positive:
$$\vec{OC} = m\left(-\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}ight) = m\left(-\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}ight)$$

### The Twist: Subtracting a Vector
The problem asks for the direction of the resultant vector $\vec{R} = \vec{OA} + \vec{OB} - \vec{OC}$.

Notice the minus sign! Subtracting a vector is physically identical to adding a vector that points in the exact opposite direction. So, $-\vec{OC}$ will flip the signs of both its components.

### Synthesizing the Resultant
Now, we simply add up all the $x$-components and $y$-components separately.

**The X-Component ($R_x$):**
$$R_x = m\left(\frac{\sqrt{3}}{2} + \frac{1}{2} - \left(-\frac{1}{\sqrt{2}}ight)ight)$$
To combine these, we use a common denominator of $2$. Since $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$, we get:
$$R_x = m\left(\frac{\sqrt{3} + 1 + \sqrt{2}}{2}ight)$$

**The Y-Component ($R_y$):**
$$R_y = m\left(\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}ight)$$
Again, using the common denominator of $2$:
$$R_y = m\left(\frac{1 - \sqrt{3} - \sqrt{2}}{2}ight)$$

### The Final Direction
To find the angle $	heta$ that this resultant vector makes with the X-axis, we take the ratio of the $y$-component to the $x$-component:
$$	an 	heta = \frac{R_y}{R_x}$$

When we divide $R_y$ by $R_x$, the magnitude $m$ and the denominator $2$ cancel out beautifully, leaving us with:
$$	an 	heta = \frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}}$$

Taking the inverse tangent gives us our final answer:
$$	heta = 	an^{-1}\left(\frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}}ight)$$

This perfectly matches option (a). By staying disciplined with our sign conventions and component resolution, a seemingly complex vector problem unravels into an elegant algebraic solution!

Animated Solution for Physics - Kinematics: The magnitude of vectors $O A$ , $O B$ , and $O C$ in the given figure are equal. The direction of $O A + O B - O C$ with X-axis will be

Select Answer:

Visualized Solution

$∣ O A ∣ = ∣ O B ∣ = ∣ O C ∣ = m$

$V = V_{x} \hat{i} + V_{y} \hat{j}$

$O A = m (\frac{3}{2} \hat{i} + \frac{1}{2} \hat{j})$

$O B = m (\frac{1}{2} \hat{i} - \frac{3}{2} \hat{j})$

$O C = m (- \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j})$

$R = O A + O B - O C$

$R_{x} = m (\frac{3 + 1 + 2}{2})$

$R_{y} = m (\frac{1 - 3 - 2}{2})$

$θ = tan^{- 1} (\frac{1 - 3 - 2}{1 + 3 + 2})$

$Conclusion$

The Sigma Insight: Vector Addition, Subtraction, and Resolution

The Visual Setup Imagine standing at the origin of a coordinate system, holding three ropes pulling in different directions

The Power of Resolution Instead of wrestling with complex geometry like the triangle or parallelogram law, we use the ultimate strategy: Vector Resolution

Breaking Down the Vectors

The Twist

Synthesizing the Resultant

The Final Direction

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Animated Solution for Physics - Kinematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The magnitude of vectors OA, OB, and OC in the given figure are equal. The direction of OA+OB−OC with X-axis will be

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣OA∣=∣OB∣=∣OC∣=m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=Vx​i^+Vy​j^​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OA=m(23​​i^+21​j^​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OB=m(21​i^−23​​j^​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OC=m(−2​1​i^+2​1​j^​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=OA+OB−OC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rx​=m(23​+1+2​​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ry​=m(21−3​−2​​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=tan−1(1+3​+2​1−3​−2​​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion