Let a and b be two vectors such that |a| = 1, |b| = 4 and a ⋅ b = 2. If c = (2a × b) - 3b and the an

Q: Wait, why is $\vec{b} \cdot (\vec{a} \times \vec{b})$ equal to zero? I thought the dot product was just multiplication.

That is a fantastic question! Remember that the cross product $(\vec{a} \times \vec{b})$ produces a vector that is perpendicular to the plane containing both $\vec{a}$ and $\vec{b}$. Since $\vec{b}$ lies in that plane, the cross product vector is perpendicular to $\vec{b}$. The dot product of any two perpendicular vectors is always zero because $\cos(90^\circ) = 0$. It is not just multiplication; it is a geometric statement about orthogonality!

Q: Can I just find the components of $\vec{a}$ and $\vec{b}$ and solve it that way?

While technically possible, it is a trap! You do not know the specific orientation of $\vec{a}$ and $\vec{b}$ in 3D space, only their magnitudes and dot product. Trying to assign arbitrary components will make the algebra unnecessarily messy and prone to errors. Stick to vector algebra properties like Lagrange's Identity; it is much more elegant and robust.

Q: Why do we calculate $|\vec{c}|^2$ instead of just finding $\vec{c}$ directly?

Because we don't have enough information to define $\vec{c}$ as a specific vector in space! We only know its relationship to $\vec{b}$ and the cross product. However, we have all the information needed to find its magnitude squared. In vector problems, when you cannot find the vector itself, always look for its magnitude or its dot product with other known vectors. That is usually where the answer hides.

Question

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = 1, |\vec{b}| = 4$ and $\vec{a} \cdot \vec{b} = 2$. If $\vec{c} = (2\vec{a} 	imes \vec{b}) - 3\vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\alpha$, then $192 \sin^2 \alpha$ is equal to _______.

Sigma Percentile · Accepted Answer

### Analyzing the Geometric Setup

We are given two vectors, $\vec{a}$ and $\vec{b}$, with defined properties. We introduce a new vector $\vec{c}$ defined by the relation:

$$\vec{c} = 2(\vec{a} 	imes \vec{b}) - 3\vec{b}$$

The cross product $(\vec{a} 	imes \vec{b})$ is geometrically perpendicular to both $\vec{a}$ and $\vec{b}$. This orthogonality is the fundamental key to simplifying the expression.

### The Dot Product Calculation

To find the angle $\alpha$ between $\vec{b}$ and $\vec{c}$, we first compute the dot product $\vec{b} \cdot \vec{c}$:

$$\vec{b} \cdot \vec{c} = \vec{b} \cdot [2(\vec{a} 	imes \vec{b}) - 3\vec{b}]$$

Expanding this, we get $2\vec{b} \cdot (\vec{a} 	imes \vec{b}) - 3(\vec{b} \cdot \vec{b})$. Because $\vec{b}$ is perpendicular to $(\vec{a} 	imes \vec{b})$, the first term vanishes:

$$\vec{b} \cdot \vec{c} = 0 - 3|\vec{b}|^2$$

Given $|\vec{b}| = 4$, we find:

$$\vec{b} \cdot \vec{c} = -3(16) = -48$$

### Determining the Magnitude of $\vec{c}$

Next, we calculate the magnitude squared, $|\vec{c}|^2$, by taking the dot product of $\vec{c}$ with itself:

$$|\vec{c}|^2 = |2(\vec{a} 	imes \vec{b}) - 3\vec{b}|^2 = 4|\vec{a} 	imes \vec{b}|^2 + 9|\vec{b}|^2$$

We apply **Lagrange's Identity** to evaluate $|\vec{a} 	imes \vec{b}|^2$:

$$|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$$

Substituting the known values $|\vec{a}|=1$, $|\vec{b}|=4$, and $\vec{a} \cdot \vec{b} = 2$:

$$|\vec{a} 	imes \vec{b}|^2 = (1)(16) - (2)^2 = 16 - 4 = 12$$

Now, substitute this back into the expression for $|\vec{c}|^2$:

$$|\vec{c}|^2 = 4(12) + 9(16) = 48 + 144 = 192$$

### Final Calculation of the Angle

We use the definition of the cosine of the angle $\alpha$ between $\vec{b}$ and $\vec{c}$:

$$\cos \alpha = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|}$$

Substituting our calculated values:

$$\cos \alpha = \frac{-48}{4 \sqrt{192}} = \frac{-12}{\sqrt{192}} = \frac{-12}{8\sqrt{3}} = -\frac{\sqrt{3}}{2}$$

Thus, $\cos^2 \alpha = \frac{3}{4}$. Finally, we determine $\sin^2 \alpha$:

$$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{3}{4} = \frac{1}{4}$$

The **final result** for the squared sine of the angle is **$1/4$**.

Animated Solution for Mathematics - Vector Algebra: Let $a$ and $b$ be two vectors such that $∣ a ∣ = 1, ∣ b ∣ = 4$ and $a \cdot b = 2$ . If $c = (2 a \times b) - 3 b$ and the angle between $b$ and $c$ is $α$ , then $192 sin^{2} α$ is equal to _______.

Enter Numerical Value:

Visualized Solution

Given Vectors $a$ and $b$

Lagrange's Identity

Calculate $∣ a \times b ∣^{2}$

Analyze Vector $c$

Dot Product $b \cdot c$

Evaluate $b \cdot c$

Calculate $∣ c ∣^{2}$

Evaluate $∣ c ∣^{2}$

Definition of $cos α$

Calculate $cos^{2} α$

Calculate $sin^{2} α$

Final Answer

The Sigma Insight: Vector (Cross) Product

Analyzing the Geometric Setup

The Dot Product Calculation

Determining the Magnitude of $c$

Final Calculation of the Angle

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Enter Numerical Value:

Visualized Solution

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