Let a, b and c be three vectors such that a = b × (b × c). If magnitudes of the vectors a, b and c a

Q: Wait, why do we use the vector triple product identity? Can't we just calculate the cross product directly?

That is a great question! While you could try to calculate the cross product directly, you don't have the components of the vectors, only their magnitudes and the angle between them. The vector triple product identity $\vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$ is specifically designed to handle this situation, allowing you to work with dot products and magnitudes instead of components. It is the most elegant and direct path to the solution.

Q: Why do we square the magnitude of the vector equation? Is that always allowed?

Yes, it is a standard and powerful technique in vector algebra! When you have an equation like $\vec{a} = \text{something}$, you can always take the dot product of both sides with themselves, which is equivalent to squaring the magnitude: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$. This is the bridge that allows us to move from the world of vectors to the world of scalars, where we can use the given magnitudes and the angle $\theta$.

Q: How do I know if $\cos \theta$ should be positive or negative?

Excellent attention to detail! The problem explicitly states that $0 < \theta < \frac{\pi}{2}$. This means $\theta$ is in the first quadrant, where all trigonometric functions, including cosine, are positive. Therefore, when you solve $\cos^2 \theta = \frac{1}{2}$, you only take the positive root, $\cos \theta = \frac{1}{\sqrt{2}}$.

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a} = \vec{b} 	imes (\vec{b} 	imes \vec{c})$. If magnitudes of the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are $\sqrt{2}, 1$ and 2 respectively and the angle between $\vec{b}$ and $\vec{c}$ is $	heta (0 < 	heta < \frac{\pi}{2})$, then the value of $1 + 	an 	heta$ is equal to:

Options:

(A) $\sqrt{3} + 1$

(B) 2

(C) 1

(D) $\frac{\sqrt{3}+1}{\sqrt{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to unravel a problem that might look intimidating at first glance, but beneath its complex exterior lies a beautiful, logical structure. We are dealing with vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, and a relationship defined by a vector triple product.

Let us walk through this step-by-step, not just to find the answer, but to understand the physics and mathematics that make it tick.

### The Triple Product Identity

We are given the relationship $\vec{a} = \vec{b} 	imes (\vec{b} 	imes \vec{c})$. This is a classic vector triple product. If you have ever felt confused by these, remember the 'BAC minus CAB' rule.

The identity states that $\vec{x} 	imes (\vec{y} 	imes \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$. Applying this to our equation, where $\vec{x} = \vec{b}$, $\vec{y} = \vec{b}$, and $\vec{z} = \vec{c}$, we get:

$$ \vec{a} = (\vec{b} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{b})\vec{c} $$

Suddenly, the complexity vanishes. We have expressed $\vec{a}$ as a linear combination of $\vec{b}$ and $\vec{c}$. This is a profound realization: it tells us that $\vec{a}$ must lie in the same plane as $\vec{b}$ and $\vec{c}$.

### Bridging to Scalar Reality

Now, we need to evaluate the dot products. We know that $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}| \cos 	heta$. Given $|\vec{b}| = 1$ and $|\vec{c}| = 2$, this becomes:

$$ \vec{b} \cdot \vec{c} = (1)(2) \cos 	heta = 2 \cos 	heta $$

Similarly, $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1^2 = 1$. Substituting these back into our expression for $\vec{a}$, we get:

$$ \vec{a} = (2 \cos 	heta)\vec{b} - \vec{c} $$

### The Power of Squaring

We are given that $|\vec{a}| = \sqrt{2}$. To utilize this, we take the magnitude squared of both sides:

$$ |\vec{a}|^2 = |2 \cos 	heta \vec{b} - \vec{c}|^2 $$

Using the property $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$, we expand the right side:

$$ |\vec{a}|^2 = (2 \cos 	heta)^2 |\vec{b}|^2 + |\vec{c}|^2 - 2(2 \cos 	heta)(\vec{b} \cdot \vec{c}) $$

Now, we substitute our known values:

$$ 2 = 4 \cos^2 	heta (1) + 4 - 4 \cos 	heta (2 \cos 	heta) $$

Simplifying this, we get $2 = 4 \cos^2 	heta + 4 - 8 \cos^2 	heta$, which further reduces to:

$$ 2 = 4 - 4 \cos^2 	heta $$

### The Trigonometric Climax

Rearranging the equation, we find $4 \cos^2 	heta = 2$, which means $\cos^2 	heta = \frac{1}{2}$. Since $	heta$ is an acute angle, $\cos 	heta = \frac{1}{\sqrt{2}}$.

This implies $	heta = \frac{\pi}{4}$. Finally, the question asks for $1 + 	an 	heta$.

Substituting $	heta = \frac{\pi}{4}$, we get:

$$ 1 + 	an\left(\frac{\pi}{4}ight) = 1 + 1 = \mathbf{2} $$

And there it is! A complex vector relationship distilled into a simple, elegant result. Keep practicing, keep visualizing, and remember that every vector problem is just a story waiting to be told.

Select Answer:

Visualized Solution

Visualizing the Vector Setup

The Vector Triple Product Identity

Applying the Identity

Evaluating the Dot Products

Substituting Dot Products

Squaring the Equation

Expanding the Squared Magnitude

Substituting Known Values

Simplifying the Equation

Solving for $cos θ$

Finding the Angle $θ$

Calculating the Final Value

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Triple Product Identity

Bridging to Scalar Reality

The Power of Squaring

The Trigonometric Climax

Similar Questions

Let $a, b$ and $c$ be three unit vectors such that $a \times (b \times c) = \frac{3}{2} (b + c)$ . If $b$ is not parallel to $c$ , then the angle between $a$ and $b$ is:

Let $a, b$ and $c$ be three non-zero vectors such that no two of them are collinear and $(a \times b) \times c = \frac{1}{3} ∣ b ∣∣ c ∣ a$ . If $θ$ is the angle between vectors $b$ and $c$ , then a value of $sin θ$ is :

Let $a, b$ and $c$ be non-zero vectors such that $(a \times b) \times c = \frac{1}{3} ∣ b ∣∣ c ∣ a$ . If $θ$ is the acute angle between the vectors $b$ and $c$ , then $sin θ$ equals

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . If $c$ is a vector such that $a \cdot c = ∣ c ∣$ , $∣ c - a ∣ = 22$ and the angle between $(a \times b)$ and $c$ is $\frac{π}{6}$ , then the value of $∣ (a \times b) \times c ∣$ is :

Let $x, y$ and $z$ be three vectors each of magnitude $2$ and the angle between each pair of them is $\frac{π}{3}$ . If $a$ is a non-zero vector perpendicular to $x$ and $y \times z$ and $b$ is a non-zero vector perpendicular to $y$ and $z \times x$ , then

If $a, b, c$ are non coplanar unit vectors such that $a \times (b \times c) = \frac{( b + c )}{2}$ , then the angle between $a$ and $b$ is

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . Let $c$ be a vector such that $∣ c - a ∣ = 3$ , $∣ (a \times b) \times c ∣ = 3$ and the angle between $c$ and $a \times b$ be $3 0^{\circ}$ . Then $a \cdot c$ is equal to:

Let $a$ , $b$ and $c$ be three vectors such that $∣ a ∣ = 3$ , $∣ b ∣ = 5$ , $b \cdot c = 10$ and angle between $b$ and $c$ is $\frac{π}{3}$ . If $a$ is perpendicular to the vector $b \times c$ , then $∣ a \times (b \times c) ∣$ is equal to

Let $a, b$ and $c$ be three vectors such that $∣ a ∣ = 3, ∣ b ∣ = 5, b . c = 10$ and the angle between $b$ and $c$ is $\frac{π}{3}$ . If $a$ is perpendicular to vector $b \times c$ , then $∣ a \times (b \times c) ∣$ is equal to ________

Let $a, b$ and $c$ be three unit vectors, out of which vectors $b$ and $c$ are non-parallel. If $α$ and $β$ are the angles which vector $a$ makes with vectors $b$ and $c$ respectively and $a \times (b \times c) = \frac{1}{2} b$ , then $∣ α - β ∣$ is equal to :