If a, b, c are non coplanar unit vectors such that a × (b × c) = (b + c)/√2, then the angle between

Q: Wait, why can we just equate the coefficients of $\vec{b}$ and $\vec{c}$? Is that always allowed?

That is a fantastic question! We can equate the coefficients because $\vec{b}$ and $\vec{c}$ are linearly independent. The problem states that $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, which implies that $\vec{b}$ and $\vec{c}$ cannot be parallel (collinear). If two vectors are not parallel, any vector in their plane can be uniquely expressed as a linear combination of them. Therefore, if $x\vec{b} + y\vec{c} = u\vec{b} + v\vec{c}$, then $x=u$ and $y=v$. It is the uniqueness of this representation that allows us to equate the coefficients.

Q: Why is the dot product $\vec{a} \cdot \vec{b}$ negative? Does that mean the angle is obtuse?

Exactly! You have hit the nail on the head. The definition of the dot product is $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. Since the magnitudes $|\vec{a}|$ and $|\vec{b}|$ are positive (they are unit vectors), the sign of the dot product depends entirely on $\cos \theta$. A negative dot product means $\cos \theta$ is negative, which occurs when the angle $\theta$ is in the second quadrant, meaning $\theta$ is obtuse (between $90^\circ$ and $180^\circ$).

Q: I am confused about the 'BAC minus CAB' rule. Does the order of the vectors matter?

Yes, the order is absolutely critical! The rule is $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. A common mistake is to write $(\vec{a} \cdot \vec{b})\vec{c} - (\vec{a} \cdot \vec{c})\vec{b}$. Always remember: the vector that is 'dotted' with the first vector (the one outside the parenthesis) is the one that gets subtracted. A helpful mnemonic is to think of the vector being 'dotted' as the one that gets 'pushed' to the end of the term.

Question

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

Options:

(A) $3\pi/4$

(B) $\pi/4$

(C) $\pi/2$

(D) $\pi$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given three non-coplanar unit vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$. The term "non-coplanar" implies that these vectors do not lie in a single flat plane and span a three-dimensional space.

Since they are unit vectors, their magnitudes are defined as $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. This property serves as the foundation for our calculations.

### The Bridge: The Vector Triple Product

We are presented with the equation:

$$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{\vec{b} + \vec{c}}{\sqrt{2}}$$

To simplify the left side, we utilize the **BAC-CAB identity**, which states that $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.

By applying this identity, we transform the complex cross product into a linear combination of $\vec{b}$ and $\vec{c}$ involving scalar dot products.

### The Power of Linear Independence

Substituting the expansion back into our original equation yields:

$$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{\sqrt{2}}\vec{b} + \frac{1}{\sqrt{2}}\vec{c}$$

Because $\vec{a}$, $\vec{b}$, and $\vec{c}$ are non-coplanar, the vectors $\vec{b}$ and $\vec{c}$ are linearly independent. This allows us to equate the coefficients of $\vec{b}$ and $\vec{c}$ on both sides of the equation.

Comparing the coefficients of $\vec{c}$, we obtain:

$$-(\vec{a} \cdot \vec{b}) = \frac{1}{\sqrt{2}}$$

This simplifies to the dot product value:

$$\vec{a} \cdot \vec{b} = -\frac{1}{\sqrt{2}}$$

### The Geometric Conclusion

We know that the dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$, where $	heta$ is the angle between the vectors. Given that $|\vec{a}| = 1$ and $|\vec{b}| = 1$, the equation becomes:

$$1 \cdot 1 \cdot \cos 	heta = -\frac{1}{\sqrt{2}}$$

This simplifies to $\cos 	heta = -\frac{1}{\sqrt{2}}$. Since the cosine function is negative in the second quadrant and the reference angle for $\frac{1}{\sqrt{2}}$ is $\frac{\pi}{4}$, we calculate:

$$	heta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$

The angle between $\vec{a}$ and $\vec{b}$ is **$\frac{3\pi}{4}$**.

Animated Solution for Mathematics - Vector Algebra: 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

Select Answer:

Visualized Solution

Given Vectors

The Given Equation

Vector Triple Product

Substituting the Expansion

Linear Independence

Equating Coefficients of $c$

Dot Product of $a$ and $b$

Dot Product Formula

Substituting Unit Magnitudes

Solving for $cos θ$

Finding the Angle $θ$

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Bridge

The Power of Linear Independence

The Geometric Conclusion

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b,c are non coplanar unit vectors such that a×(b×c)=2​(b+c)​, then the angle between a and b is