Let the vectors a, b, c and d be such that (a x b) x (c x d) = 0. Let P1 and P2 be planes determined

Q: Wait, why do we use the cross product to find the normal vector? Can't we just use the dot product?

Great question! The dot product of two vectors gives us a scalar, which tells us about their projection or angle, but it doesn't give us a new vector. The cross product, however, is specifically designed to produce a vector that is perpendicular to both input vectors. Since a plane is defined by two vectors, their cross product naturally points in the direction perpendicular to that plane, making it the perfect tool for finding the normal vector.

Q: Does the equation $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$ mean that the vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are all zero?

Not at all! The equation tells us that the *result* of the cross product of the two normal vectors is the zero vector. It doesn't mean the individual vectors are zero. It simply means the two normal vectors are parallel to each other. Think of it like two flagpoles: they don't have to be zero-length to be parallel; they just need to point in the same (or opposite) direction.

Q: If the normal vectors are parallel, why is the angle between the planes $0$?

Think of the normal vector as the 'direction' the plane is facing. If two planes have parallel normal vectors, they are facing the exact same direction in space. If they are facing the same direction, they are parallel to each other. By definition, the angle between two parallel planes is $0$ degrees (or $0$ radians). It's a beautiful geometric consequence!

Question

Let the vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ be such that $(\vec{a} 	imes \vec{b}) 	imes (\vec{c} 	imes \vec{d}) = \vec{0}$. Let $P_1$ and $P_2$ be planes determined by the pairs of vectors $\vec{a}, \vec{b}$ and $\vec{c}, \vec{d}$ respectively. Then the angle between $P_1$ and $P_2$ is

Options:

(A) $0$

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

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

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

Sigma Percentile · Accepted Answer

### The Geometry of Orientation: A Journey into 3D Space

Imagine you are standing in a vast, empty three-dimensional room. You have two flat sheets of paper, $P_1$ and $P_2$, floating in the air.

These aren't just random sheets; they are anchored by vectors. Plane $P_1$ is defined by vectors $\vec{a}$ and $\vec{b}$, which lie perfectly flat on its surface. Similarly, plane $P_2$ is defined by vectors $\vec{c}$ and $\vec{d}$.

These vectors are the DNA of the planes, dictating exactly how they are tilted and positioned in space.

### The Flagpole Analogy

To understand how these planes are oriented, we need a reference. In geometry, we use the normal vector.

Imagine placing a flagpole perpendicular to each sheet of paper. For plane $P_1$, this flagpole is the vector $\vec{n}_1 = \vec{a} 	imes \vec{b}$.

This vector shoots straight out from the surface, perpendicular to every line on the plane. It is the 'soul' of the plane's orientation. We do the same for plane $P_2$, creating a second flagpole, $\vec{n}_2 = \vec{c} 	imes \vec{d}$.

Now, we have two planes and two flagpoles, and we are ready to tackle the problem's core.

### Decoding the Intimidating Equation

The problem presents us with a seemingly complex condition:

$$(\vec{a} 	imes \vec{b}) 	imes (\vec{c} 	imes \vec{d}) = \vec{0}$$

At first glance, this looks like a mess of brackets and cross products. But let's breathe and look closer.

We have already defined our normal vectors! The term $(\vec{a} 	imes \vec{b})$ is just $\vec{n}_1$, and $(\vec{c} 	imes \vec{d})$ is just $\vec{n}_2$.

Substituting these in, the equation simplifies to something incredibly elegant:

$$\vec{n}_1 	imes \vec{n}_2 = \vec{0}$$

### The Revelation of Parallelism

Now, we must ask: what does it mean for the cross product of two vectors to be the zero vector?

Geometrically, the magnitude of the cross product is given by $|\vec{n}_1| |\vec{n}_2| \sin(	heta)$, where $	heta$ is the angle between the vectors. For this to be zero, $\sin(	heta)$ must be zero, which means $	heta$ must be $0$ or $\pi$.

This happens if and only if the vectors are parallel or collinear. Therefore, our two flagpoles, $\vec{n}_1$ and $\vec{n}_2$, are parallel.

### The Final Synthesis

If the normal vectors, which define the orientation of the planes, are parallel, then the planes themselves must be parallel.

It is a beautiful, logical chain: the vectors define the planes, the cross product defines the normals, the equation defines the relationship between the normals, and that relationship defines the angle between the planes.

Since the planes are parallel, the angle between them is **$0$**. This problem isn't just about calculation; it's about seeing the hidden geometric reality behind the algebraic symbols. You've just mastered the art of 3D orientation!

Animated Solution for Mathematics - Vector Algebra: Let the vectors $a, b, c$ and $d$ be such that $(a \times b) \times (c \times d) = 0$ . Let $P_{1}$ and $P_{2}$ be planes determined by the pairs of vectors $a, b$ and $c, d$ respectively. Then the angle between $P_{1}$ and $P_{2}$ is

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Visualized Solution

Plane $P_{1}$ and its Vectors

Normal to Plane $P_{1}$

Plane $P_{2}$ and its Vectors

Normal to Plane $P_{2}$

The Given Condition

Substituting the Normals

Interpreting the Cross Product

Angle Between the Planes

The Sigma Insight: Vector (Cross) Product

The Geometry of Orientation

The Flagpole Analogy

Decoding the Intimidating Equation

The Revelation of Parallelism

The Final Synthesis

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Visualized Solution

Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​ and its Vectors

Normal to Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​

Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​ and its Vectors

Normal to Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​