If a and b are two unit vectors such that a+2b and 5a-4b are perpendicular to each other then the an

Q: Wait, why can we just distribute the dot product like normal multiplication?

Great question! The dot product follows the distributive law over vector addition, just like standard multiplication. Because it maps vectors to a scalar, the algebraic rules of expansion hold perfectly. You are essentially projecting each component of the first vector onto the second, and the linearity of the projection ensures the distributive property remains valid.

Q: Can I just assume the vectors are along the x and y axes to solve this?

While that might work for a specific case, it is a dangerous habit! In JEE Advanced, you must solve for the general case. By using the dot product definition $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$, you ensure your result holds true for any orientation of $\vec{a}$ and $\vec{b}$ in space.

Q: Why does $\vec{a} \cdot \vec{b}$ become $\vec{b} \cdot \vec{a}$?

That is the commutative property of the dot product. Since the dot product is defined as $|\vec{a}||\vec{b}| \cos \theta$, and the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is the same as the angle between $\vec{b}$ and $\vec{a}$, the order of the vectors does not change the scalar result. It is a beautiful symmetry of Euclidean space!

Q: What if the vectors weren't unit vectors? Would the answer change?

Absolutely! If they weren't unit vectors, we would need to know their magnitudes to solve for $\cos \theta$. The 'unit vector' condition is a gift—it simplifies the equation by setting $|\vec{a}|^2$ and $|\vec{b}|^2$ to 1, allowing us to isolate the cosine term directly.

Q: Is there a geometric way to visualize this without all the algebra?

You could think of this as finding the specific orientation where the 'weighted sum' of these vectors results in a zero projection. However, the algebra is the language of physics here—it is the most robust way to ensure you don't miss any subtle geometric constraints.

Question

If $\vec{a}$ and $\vec{b}$ are two unit vectors such that $\vec{a}+2\vec{b}$ and $5\vec{a}-4\vec{b}$ are perpendicular to each other then the angle between $\vec{a}$ and $\vec{b}$ is

Options:

(A) $45^\circ$

(B) $60^\circ$

(C) $\cos^{-1}\left(\frac{1}{3}\right)$

(D) $\cos^{-1}\left(\frac{2}{7}\right)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty space. You have two unit vectors, $\vec{a}$ and $\vec{b}$, stretching out from your position like two needles on a compass.

We are given two new vectors, $\vec{u} = \vec{a} + 2\vec{b}$ and $\vec{v} = 5\vec{a} - 4\vec{b}$. We are told that these two vectors are perfectly perpendicular, which serves as the geometric key to solving the problem.

### The Power of the Dot Product

When we encounter the condition of perpendicularity, we must utilize the **dot product**. If two vectors are perpendicular, their dot product must be zero because $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos(90^{\circ}) = 0$.

This geometric condition translates into the algebraic equation:
$$\vec{u} \cdot \vec{v} = 0$$

### Expanding the Horizon

We substitute the definitions of $\vec{u}$ and $\vec{v}$ into the dot product equation:
$$(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0$$

By distributing the dot product across the terms, we expand the expression:
$$\vec{a} \cdot (5\vec{a}) + \vec{a} \cdot (-4\vec{b}) + (2\vec{b}) \cdot (5\vec{a}) + (2\vec{b}) \cdot (-4\vec{b}) = 0$$

### The Beauty of Symmetry

Simplifying the expression using the scalar properties of dot products, we obtain:
$$5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0$$

Since the dot product is commutative ($\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b}$), we combine the middle terms:
$$5(\vec{a} \cdot \vec{a}) + 6(\vec{a} \cdot \vec{b}) - 8(\vec{b} \cdot \vec{b}) = 0$$

Because $\vec{a}$ and $\vec{b}$ are unit vectors, we know that $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1$. Substituting these values yields:
$$5(1) + 6(\vec{a} \cdot \vec{b}) - 8(1) = 0$$

### The Final Revelation

Simplifying the linear expression, we find:
$$-3 + 6(\vec{a} \cdot \vec{b}) = 0 \Rightarrow \vec{a} \cdot \vec{b} = \frac{3}{6} = \frac{1}{2}$$

Recalling the definition $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta$, and knowing $|\vec{a}| = 1$ and $|\vec{b}| = 1$, we have:
$$\cos 	heta = \frac{1}{2}$$

The angle $	heta$ that satisfies this condition is **$60^{\circ}$** (or $\frac{\pi}{3}$ radians).

Animated Solution for Mathematics - Vector Algebra: If $a$ and $b$ are two unit vectors such that $a + 2 b$ and $5 a - 4 b$ are perpendicular to each other then the angle between $a$ and $b$ is

Select Answer:

Visualized Solution

Defining the Base Vectors

Constructing the Linear Combinations

The Perpendicularity Condition

Setting Up the Dot Product Equation

Expanding the Dot Product

Grouping Like Terms

Applying Commutativity

Using Unit Vector Properties

Simplifying the Equation

Solving for the Dot Product

Finding the Angle $θ$

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Power of the Dot Product

Expanding the Horizon

The Beauty of Symmetry

The Final Revelation

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