Let a and b be two unit vectors such that |(a+b) + 2(a × b)| = 2. If θ ∈ (0, π) is the angle between

Q: Why do we square the magnitude? Can't we just solve it directly?

Great question! In vector algebra, the magnitude operator $|\cdot|$ is non-linear and difficult to manipulate directly. Squaring both sides is the standard 'vector-killer' technique. It transforms the magnitude into a dot product, specifically $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$, which allows us to use the distributive property and simplify the expression using known identities like $\hat{a} \cdot \hat{a} = 1$ and $\hat{a} \cdot \hat{b} = \cos\theta$.

Q: Why is the dot product of $(\hat{a}+\hat{b})$ and $(\hat{a} \times \hat{b})$ zero?

This is the core geometric insight of the problem! The cross product $\hat{a} \times \hat{b}$ produces a vector that is, by definition, perpendicular to the plane containing both $\hat{a}$ and $\hat{b}$. Since the sum vector $(\hat{a}+\hat{b})$ also lies within that same plane, the cross product vector must be perpendicular to the sum vector. The dot product of any two perpendicular vectors is always zero.

Q: Why do we reject the solution $\cos\theta = 1$?

The problem explicitly constrains the angle $\theta$ to the open interval $(0, \pi)$. If $\cos\theta = 1$, then $\theta = 0$. Since $0$ is not in the open interval $(0, \pi)$, we must discard this solution. Always keep a sharp eye on the domain constraints provided in the question; they are often the key to filtering out extraneous roots.

Question

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that $|(\hat{a}+\hat{b}) + 2(\hat{a} 	imes \hat{b})| = 2$. If $	heta \in (0, \pi)$ is the angle between $\hat{a}$ and $\hat{b}$, then among the statements :
(S1) : $2|\hat{a} 	imes \hat{b}| = |\hat{a} - \hat{b}|$
(S2) : The projection of $\hat{a}$ on $(\hat{a} + \hat{b})$ is $\frac{1}{2}$

Options:

(A) Only (S1) is true

(B) Only (S2) is true

(C) Both (S1) and (S2) are true

(D) Both (S1) and (S2) are false

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dissect a problem that perfectly encapsulates the elegance of vector algebra. It is not just about crunching numbers; it is about visualizing the hidden relationships between vectors.

We are given two unit vectors, $\hat{a}$ and $\hat{b}$, with an angle $	heta$ between them, and a mysterious equation:
$$|(\hat{a}+\hat{b}) + 2(\hat{a} 	imes \hat{b})| = 2$$

At first glance, this looks like a chaotic mix of sums and cross products. But fear not! We are going to peel back the layers.

### The Power of Squaring

When you face a magnitude equation, your first instinct should always be to square it. The magnitude operator is a bit of a black box, but the dot product is a transparent, linear tool.

By squaring both sides, we transform the equation into:
$$|(\hat{a}+\hat{b}) + 2(\hat{a} 	imes \hat{b})|^2 = 4$$

Now, we use the expansion rule $|\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$. Here, our $\vec{x}$ is $(\hat{a}+\hat{b})$ and our $\vec{y}$ is $2(\hat{a} 	imes \hat{b})$. This gives us three distinct terms: the magnitude of the sum squared, the magnitude of the cross product squared, and the cross-term dot product.

### The Geometric Insight

This is where the magic happens. Look at that cross-term: $2(\hat{a}+\hat{b}) \cdot (2(\hat{a} 	imes \hat{b}))$.

We know that $\hat{a} 	imes \hat{b}$ is a vector perpendicular to the plane of $\hat{a}$ and $\hat{b}$. The sum $(\hat{a}+\hat{b})$ lies entirely within that same plane. Therefore, the dot product of a vector in the plane and a vector perpendicular to the plane is zero!

The entire cross-term vanishes into thin air. We are left with a much cleaner equation:
$$|\hat{a}+\hat{b}|^2 + 4|\hat{a} 	imes \hat{b}|^2 = 4$$

### The Trigonometric Bridge

Now, we translate this geometry into trigonometry. We know that $|\hat{a}+\hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 + 2|\hat{a}||\hat{b}|\cos	heta$. Since they are unit vectors, this simplifies to $2 + 2\cos	heta$.

Similarly, the magnitude of the cross product is $|\hat{a}||\hat{b}|\sin	heta$, so its square is $\sin^2	heta$. Substituting these into our simplified equation, we get:
$$(2 + 2\cos	heta) + 4\sin^2	heta = 4$$

By converting $\sin^2	heta$ to $1 - \cos^2	heta$, we arrive at the quadratic equation:
$$2\cos^2	heta - \cos	heta - 1 = 0$$

Solving this, we find $\cos	heta = -\frac{1}{2}$ or $\cos	heta = 1$. Given our constraint $	heta \in (0, \pi)$, we reject $\cos	heta = 1$ and accept $\cos	heta = -\frac{1}{2}$, which means $	heta = \frac{2\pi}{3}$.

### Verification and Victory

With $	heta = \frac{2\pi}{3}$ in hand, we check our statements. For (S1), we calculate $2|\hat{a} 	imes \hat{b}| = 2\sin(\frac{2\pi}{3}) = \sqrt{3}$, and $|\hat{a} - \hat{b}| = \sqrt{1 + 1 - 2\cos(\frac{2\pi}{3})} = \sqrt{3}$. They match!

For (S2), the projection of $\hat{a}$ on $(\hat{a} + \hat{b})$ is:
$$\frac{\hat{a} \cdot (\hat{a} + \hat{b})}{|\hat{a} + \hat{b}|} = \frac{1 + \cos	heta}{\sqrt{2 + 2\cos	heta}} = \frac{1 - 1/2}{1} = \frac{1}{2}$$

**Both statements are true!** You have successfully navigated the vector landscape. Keep this confidence; you are ready for any challenge.

Select Answer:

Visualized Solution

Visualizing the Vectors $\overset{a}{^}$ and $\hat{b}$

Squaring the Magnitude Equation

Expanding the Equation

Applying Orthogonality Property

Expanding in terms of $θ$

Forming the Quadratic Equation

Solving for $cos θ$

Finding the Angle $θ$

Checking Statement (S1) - Part 1

Checking Statement (S1) - Part 2

Checking Statement (S2) - Projection Formula

Calculating the Projection

Final Conclusion

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Power of Squaring

The Geometric Insight

The Trigonometric Bridge

Verification and Victory

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