Let a and b be two unit vectors. If the vectors c = a + 2b and d = 5a - 4b are perpendicular to each

Q: Wait, why is the dot product zero? Can't we just use the magnitude formula?

Great question! The magnitude formula relates to the length of a vector, but the condition of perpendicularity is specifically about the *angle* between them. The dot product $\vec{c} \cdot \vec{d} = |\vec{c}||\vec{d}| \cos(\theta)$ is the most direct way to use the $90^\circ$ angle, because $\cos(90^\circ) = 0$. It is the most efficient path to the solution.

Q: Why does $\vec{a} \cdot \vec{a} = 1$? I thought it was just a vector.

Remember, $\vec{a}$ is a unit vector. By definition, a unit vector has a magnitude of $1$. The dot product of any vector with itself is the square of its magnitude: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$. Since $|\vec{a}| = 1$, then $1^2 = 1$. It is a powerful property that simplifies complex expressions!

Q: Why can't we just add the vectors directly?

You could, but that would give you the resultant vector, not the angle between the original vectors. The problem asks for the angle between $\vec{a}$ and $\vec{b}$, which is hidden inside the dot product $\vec{a} \cdot \vec{b}$. Expanding the dot product is the algebraic way to extract that information.

Question

Let $\vec{a}$ and $\vec{b}$ be two unit vectors. If the vectors $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d} = 5\vec{a} - 4\vec{b}$ are perpendicular to each other, then the angle between $\vec{a}$ and $\vec{b}$ is:

Options:

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{2}$

(C) $\frac{\pi}{3}$

(D) $\frac{\pi}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, traveler of the mathematical realms! Today, we are going to unravel a beautiful problem that sits at the heart of vector algebra. We are given two unit vectors, $\vec{a}$ and $\vec{b}$, and we need to find the angle between them.

Imagine you are standing in a coordinate plane. You have two unit vectors, $\vec{a}$ and $\vec{b}$, both starting from the origin and stretching out to a distance of exactly $1$. They are like two hands of a clock, and our goal is to find the angle $	heta$ between them.

We are given two new vectors, $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d} = 5\vec{a} - 4\vec{b}$, and we are told they are perpendicular. This is our golden ticket.

### The Power of the Dot Product

When we say two vectors are perpendicular, or orthogonal, we are saying they meet at a $90^{\circ}$ angle. In the language of vectors, this is a profound statement.

The dot product of two vectors is defined as $\vec{c} \cdot \vec{d} = |\vec{c}||\vec{d}| \cos(	heta)$. Since the angle between $\vec{c}$ and $\vec{d}$ is $90^{\circ}$, and $\cos(90^{\circ}) = 0$, the dot product must be zero.

This is the bridge that connects the geometry of the problem to the algebra we need to perform. So, we write:
$$\vec{c} \cdot \vec{d} = 0$$

Now, let us substitute the expressions for $\vec{c}$ and $\vec{d}$ into this equation:
$$(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0$$

### The Algebraic Expansion

Let us expand this carefully. We distribute the dot product across the terms:
$$\vec{a} \cdot (5\vec{a} - 4\vec{b}) + 2\vec{b} \cdot (5\vec{a} - 4\vec{b}) = 0$$

This gives us four distinct terms:
$$5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0$$

We know that the dot product is commutative, so $\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b}$. This allows us to combine the middle terms:
$$-4(\vec{a} \cdot \vec{b}) + 10(\vec{a} \cdot \vec{b}) = 6(\vec{a} \cdot \vec{b})$$

Now, our equation simplifies to:
$$5(\vec{a} \cdot \vec{a}) + 6(\vec{a} \cdot \vec{b}) - 8(\vec{b} \cdot \vec{b}) = 0$$

### The Elegance of Unit Vectors

Here is where the magic happens. We are given that $\vec{a}$ and $\vec{b}$ are unit vectors, meaning $|\vec{a}| = 1$ and $|\vec{b}| = 1$. The dot product of a vector with itself is the square of its magnitude:
$$\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1^2 = 1$$
$$\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1^2 = 1$$

Substituting these values into our equation, we get:
$$5(1) + 6(\vec{a} \cdot \vec{b}) - 8(1) = 0$$

This simplifies to:
$$6(\vec{a} \cdot \vec{b}) - 3 = 0$$

Solving for the dot product, we find:
$$\vec{a} \cdot \vec{b} = \frac{3}{6} = \frac{1}{2}$$

### The Final Reveal

We have reached the final step of our journey. We know that $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(	heta)$. Since both are unit vectors, $|\vec{a}| = 1$ and $|\vec{b}| = 1$, so $\vec{a} \cdot \vec{b} = \cos(	heta)$.

We just found that $\vec{a} \cdot \vec{b} = \frac{1}{2}$, so:
$$\cos(	heta) = \frac{1}{2}$$

The angle whose cosine is $\frac{1}{2}$ is $\frac{\pi}{3}$, or $60^{\circ}$. We have successfully navigated the algebra and arrived at the **final answer**:
$$	heta = 60^{\circ} 	ext{ or } \frac{\pi}{3}$$

Animated Solution for Mathematics - Vector Algebra: Let $a$ and $b$ be two unit vectors. If the vectors $c = a + 2 b$ and $d = 5 a - 4 b$ are perpendicular to each other, then the angle between $a$ and $b$ is:

Select Answer:

Visualized Solution

Visualizing the Unit Vectors $a$ and $b$

The Condition for Perpendicularity

Substituting the Vector Expressions

Expanding the Dot Product

Applying Unit Vector Properties

Simplifying the Algebraic 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

The Algebraic Expansion

The Elegance of Unit Vectors

The Final Reveal

Similar Questions

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

If $(a + 3 b)$ is perpendicular to $(7 a - 5 b)$ and $(a - 4 b)$ is perpendicular to $(7 a - 2 b)$ , then the angle between $a$ and $b$ (in degrees) is ___

Let $a$ and $b$ be two vectors such that $∣2 a + 3 b ∣ = ∣3 a + b ∣$ and the angle between $a$ and $b$ is $6 0^{\circ}$ . If $\frac{1}{8} a$ is a unit vector, then $∣ b ∣$ is equal to :

If $a, b, c$ are vectors show that $a + b + c = 0$ and $∣ a ∣ = 7, ∣ b ∣ = 5, ∣ c ∣ = 3$ then angle between vector $b$ and $c$ is

Let $a, b, c \in R$ be such that $a^{2} + b^{2} + c^{2} = 1$ . If $a cos θ = b cos (θ + \frac{2 π}{3}) = c cos (θ + \frac{4 π}{3})$ , where $θ = \frac{π}{9}$ , then the angle between the vectors $a \hat{i} + b \hat{j} + c \hat{k}$ and $b \hat{i} + c \hat{j} + a \hat{k}$ is :

The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

Let the angle $θ, 0 < θ < \frac{π}{2}$ between two unit vectors $\overset{a}{^}$ and $\hat{b}$ be $sin^{- 1} (\frac{65}{9})$ . If the vector $c = 3 \overset{a}{^} + 6 \hat{b} + 9 (\overset{a}{^} \times \hat{b})$ , then the value of $9 (c \cdot \overset{a}{^}) - 3 (c \cdot \hat{b})$ is

Let the vectors $a, b, c$ be such that $∣ a ∣ = 2, ∣ b ∣ = 4$ and $∣ c ∣ = 4$ . If the projection of $b$ on $a$ is equal to the projection of $c$ on $a$ and $b$ is perpendicular to $c$ , then the value of $∣ a + b - c ∣$ is

Let $A, B, C$ be vectors of length 3, 4, 5 respectively. Let $A$ be perpendicular to $B + C$ , $B$ to $C + A$ and $C$ to $A + B$ . Then the length of vector $A + B + C$ is .........