If a, b and c are unit vectors such that a + 2b + 2c = 0, then |a × c| is equal to :
(1) 1/4
(2) √15

Q: Why do we square the equation? Can't we just use the triangle law?

The triangle law is excellent for visualization, but it doesn't directly give you the numerical value of the cross product. Squaring the equation is the algebraic bridge that converts vector relationships into scalar dot products. The dot product is the mathematical tool that links the vectors to the cosine of the angle between them, which is exactly what we need to find the sine for the cross product.

Q: Wait, why do we isolate $2\vec{b}$ specifically? Why not isolate $\vec{a}$?

You absolutely could isolate $\vec{a}$, but that would leave you with $\vec{b}$ and $\vec{c}$ on the other side. Squaring that would give you the dot product $\vec{b} \cdot \vec{c}$, which doesn't help us find $|\vec{a} \times \vec{c}|$. We isolate the vector we *don't* need to find the dot product of the two we *do* need.

Q: Is the negative sign in $\vec{a} \cdot \vec{c} = -\frac{1}{4}$ a problem for the cross product?

Not at all! A negative dot product simply tells us that the angle between $\vec{a}$ and $\vec{c}$ is obtuse (greater than $90^\circ$). The magnitude of the cross product, $|\vec{a} \times \vec{c}| = |\vec{a}||\vec{c}| \sin \theta$, depends on $\sin \theta$. Since $\sin \theta$ is always positive for any angle between $0$ and $\pi$, the final magnitude will always be positive, regardless of the sign of the dot product.

Question

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a} + 2\vec{b} + 2\vec{c} = 0$, then $|\vec{a} 	imes \vec{c}|$ is equal to :

Options:

(A) 1/4

(B) $\sqrt{15}/16$

(C) 15/16

(D) $\sqrt{15}/4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional space, holding three arrows of equal length—unit vectors $\vec{a}, \vec{b},$ and $\vec{c}$. They are bound by a rigid constraint: $\vec{a} + 2\vec{b} + 2\vec{c} = 0$.

This equation implies that if you place these vectors head-to-tail, they form a closed loop. In the context of JEE Advanced, we must extract the hidden relationships within this geometry.

Our mission is to find the magnitude of the cross product $|\vec{a} 	imes \vec{c}|$. Since $|\vec{a} 	imes \vec{c}| = |\vec{a}||\vec{c}| \sin 	heta$ and the vectors are unit vectors ($|\vec{a}|=|\vec{c}|=1$), the problem reduces to finding $\sin 	heta$, where $	heta$ is the angle between $\vec{a}$ and $\vec{c}$.

### The Algebraic Strategy: Squaring to Success

To find $	heta$, we must determine the dot product $\vec{a} \cdot \vec{c}$. We utilize the identity $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$ starting from our constraint $\vec{a} + 2\vec{b} + 2\vec{c} = 0$.

To isolate the relationship between $\vec{a}$ and $\vec{c}$, we rearrange the terms to isolate $\vec{b}$:

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

Now, we square both sides. Because $|-\vec{v}|^2 = |\vec{v}|^2$, the negative sign vanishes:

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

$$4|\vec{b}|^2 = |\vec{a} + 2\vec{c}|^2$$

### Expanding the Soul of the Equation

We expand the right side using the identity $|\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$. Applying this to $|\vec{a} + 2\vec{c}|^2$, we obtain:

$$4|\vec{b}|^2 = |\vec{a}|^2 + |2\vec{c}|^2 + 2(\vec{a} \cdot 2\vec{c})$$

$$4|\vec{b}|^2 = |\vec{a}|^2 + 4|\vec{c}|^2 + 4(\vec{a} \cdot \vec{c})$$

Since $\vec{a}, \vec{b},$ and $\vec{c}$ are unit vectors, their magnitudes are $1$. Substituting these values yields:

$$4(1)^2 = (1)^2 + 4(1)^2 + 4(\vec{a} \cdot \vec{c})$$

$$4 = 5 + 4(\vec{a} \cdot \vec{c})$$

Solving for the dot product, we find:

$$-1 = 4(\vec{a} \cdot \vec{c}) \ \Rightarrow \vec{a} \cdot \vec{c} = -\frac{1}{4}$$

### The Final Bridge: From Dot Product to Cross Product

We know that $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos 	heta$. Given the unit magnitudes, we have $\cos 	heta = -\frac{1}{4}$.

We now use the trigonometric identity $\sin^2 	heta = 1 - \cos^2 	heta$ to find $\sin 	heta$:

$$\sin^2 	heta = 1 - \left(-\frac{1}{4}ight)^2$$

$$\sin^2 	heta = 1 - \frac{1}{16} = \frac{15}{16}$$

$$\sin 	heta = \frac{\sqrt{15}}{4}$$

Finally, the magnitude of the cross product is:

$$|\vec{a} 	imes \vec{c}| = |\vec{a}||\vec{c}| \sin 	heta = (1)(1)\left(\frac{\sqrt{15}}{4}ight) = \mathbf{\frac{\sqrt{15}}{4}}$$

Animated Solution for Mathematics - Vector Algebra: If $a, b$ and $c$ are unit vectors such that $a + 2 b + 2 c = 0$ , then $∣ a \times c ∣$ is equal to :

Select Answer:

Visualized Solution

Identify Unit Vectors

Isolating the Target Vector

Squaring Both Sides

Expanding the Right Side

Substituting Unit Magnitudes

Solving for the Dot Product

Dot Product to Cosine

Finding Sine of the Angle

Calculating the Cross Product

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Algebraic Strategy

Expanding the Soul of the Equation

The Final Bridge

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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 and c are unit vectors such that a+2b+2c=0, then ∣a×c∣ is equal to :