If A, B, C are three non-coplanar vectors, then (A.B x C)/(C x A.B) + (B.A x C)/(C.A x B) = ........

Q: Wait, why can we just move the dot and cross around in the Scalar Triple Product?

Think of the Scalar Triple Product as the volume of a parallelepiped. Whether you calculate the base area first (using the cross product) and then find the height (using the dot product), or vice versa, the volume remains the same! The dot and cross are just different ways to slice that volume, and the cyclic property ensures the result is invariant.

Q: Why does swapping two vectors make the whole thing negative?

Because the cross product is anti-commutative! $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$. When you swap vectors inside the box product, you are essentially flipping the direction of the cross product, which naturally flips the sign of the entire volume. It is a direct consequence of the right-hand rule.

Q: What happens if the vectors are coplanar?

If they are coplanar, they lie on a flat plane, meaning the volume of the parallelepiped is zero. In that case, the denominator would be zero, and the expression would be undefined. The problem specifically states they are non-coplanar to ensure we are working with a valid, non-zero volume.

Question

If $\vec{A}, \vec{B}, \vec{C}$ are three non-coplanar vectors, then $\frac{\vec{A} \cdot \vec{B} 	imes \vec{C}}{\vec{C} 	imes \vec{A} \cdot \vec{B}} + \frac{\vec{B} \cdot \vec{A} 	imes \vec{C}}{\vec{C} \cdot \vec{A} 	imes \vec{B}} = .........$

Sigma Percentile · Accepted Answer

### The Geometry of Space

Imagine you are standing in a vast, empty room. You have three vectors, $\vec{A}, \vec{B},$ and $\vec{C}$, all originating from the same point.

Because they are non-coplanar, they do not lie flat on a single sheet of paper. Instead, they stretch out into three-dimensional space, forming the edges of a slanted 3D box—a parallelepiped.

The volume of this shape is the physical soul of the Scalar Triple Product (STP), denoted as $[\vec{A} \vec{B} \vec{C}]$. Since the problem guarantees they are non-coplanar, we know this volume is non-zero, which is our safety net against division by zero.

### The Cyclic Dance

In the world of JEE vector algebra, the Scalar Triple Product is your most elegant weapon. It is defined as:

$$[\vec{A} \vec{B} \vec{C}] = \vec{A} \cdot (\vec{B} 	imes \vec{C})$$

A beautiful property of the STP is that the dot and cross operations can be interchanged without changing the result:

$$[\vec{A} \vec{B} \vec{C}] = (\vec{A} 	imes \vec{B}) \cdot \vec{C}$$

Even more powerful is the cyclic property: if we shift the vectors in a cyclic order—moving $\vec{A}$ to the end to get $\vec{B}, \vec{C}, \vec{A}$—the value remains completely unchanged. Thus:

$$[\vec{A} \vec{B} \vec{C}] = [\vec{B} \vec{C} \vec{A}] = [\vec{C} \vec{A} \vec{B}]$$

This is the key to unlocking our expression.

### The First Term

Let us look at the first term:

$$\frac{\vec{A} \cdot (\vec{B} 	imes \vec{C})}{(\vec{C} 	imes \vec{A}) \cdot \vec{B}}$$

The numerator is the standard box product $[\vec{A} \vec{B} \vec{C}]$. Now, look at the denominator: $(\vec{C} 	imes \vec{A}) \cdot \vec{B}$.

Using the commutativity of the dot product, this is $\vec{B} \cdot (\vec{C} 	imes \vec{A})$, which is the box product $[\vec{B} \vec{C} \vec{A}]$. By our cyclic property, $[\vec{B} \vec{C} \vec{A}]$ is exactly equal to $[\vec{A} \vec{B} \vec{C}]$.

Thus, the first term is simply:

$$\frac{[\vec{A} \vec{B} \vec{C}]}{[\vec{A} \vec{B} \vec{C}]} = 1$$

It collapses beautifully.

### The Anti-Symmetry Trap

Now, we face the second term:

$$\frac{\vec{B} \cdot (\vec{A} 	imes \vec{C})}{\vec{C} \cdot (\vec{A} 	imes \vec{B})}$$

The numerator is $\vec{B} \cdot (\vec{A} 	imes \vec{C}) = [\vec{B} \vec{A} \vec{C}]$. The denominator is $\vec{C} \cdot (\vec{A} 	imes \vec{B}) = [\vec{C} \vec{A} \vec{B}]$.

We need to convert these to our standard $[\vec{A} \vec{B} \vec{C}]$. For the numerator, we use the anti-symmetry property: swapping $\vec{A}$ and $\vec{B}$ in $[\vec{B} \vec{A} \vec{C}]$ gives us $-[\vec{A} \vec{B} \vec{C}]$.

For the denominator, $[\vec{C} \vec{A} \vec{B}]$ is just a cyclic shift of $[\vec{A} \vec{B} \vec{C}]$, so it remains $[\vec{A} \vec{B} \vec{C}]$.

### The Final Collapse

Substituting these back, the second term becomes:

$$\frac{-[\vec{A} \vec{B} \vec{C}]}{[\vec{A} \vec{B} \vec{C}]} = -1$$

We are at the finish line. The first term gave us $1$, and the second term gave us $-1$.

Adding them together, $1 + (-1) = 0$. The entire complex-looking vector expression vanishes into **zero**.

This is the elegance of vector algebra—what looks like a mountain of complexity is often just a perfectly balanced symmetry waiting to be revealed.

Animated Solution for Mathematics - Vector Algebra: If $A, B, C$ are three non-coplanar vectors, then $\frac{A \cdot B \times C}{C \times A \cdot B} + \frac{B \cdot A \times C}{C \cdot A \times B} = .........$

Enter Numerical Value:

Visualized Solution

Visualizing Non-coplanar Vectors

Defining the Scalar Triple Product

The Cyclic Property of STP

Analyzing the First Term

Simplifying the First Term

The Anti-symmetry Property

Analyzing the Second Term

Simplifying the Second Term

Final Summation

Conclusion & Key Takeaways

The Sigma Insight: Scalar Triple Product

The Geometry of Space

The Cyclic Dance

The First Term

The Anti-Symmetry Trap

The Final Collapse

Similar Questions

If $u, v$ and $w$ are three non-coplanar vectors, then $(u + v - w) \cdot [(u - v) \times (v - w)]$ equals

If $a, b$ and $c$ are three non coplanar vectors, then $(a + b + c) \cdot [(a + b) \times (a + c)]$ equals

Let $a, b, c$ be three non-coplanar vectors and $p, q, r$ are vectors defined by the relations $p = \frac{b \times c}{[ a b c ]}, q = \frac{c \times a}{[ a b c ]}, r = \frac{a \times b}{[ a b c ]}$ then the value of the expression $(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r$ is equal to

If $a, b, c$ are non coplanar vectors and $λ$ is a real number then $[λ (a + b) λ^{2} b λ c] = [a b + c b]$ for

If vectors $a, b, c$ are coplanar, show that $a \cdot a b \cdot a c \cdot a a \cdot b b \cdot b c \cdot b a \cdot c b \cdot c c \cdot c = 0$ .

If $a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $a \cdot c = 5, b ⊥ c$ , then $122 (c_{1} + c_{2} + c_{3})$ is equal to

If the vectors $a \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k}$ ( $a \neq = b \neq = c \neq = 1$ ) are coplanar, then the value of $\frac{1}{( 1 - a )} + \frac{1}{( 1 - b )} + \frac{1}{( 1 - c )} = .........$

If four distinct points with position vectors $a, b, c$ and $d$ are coplanar, then $[a b c]$ is equal to

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 three non-coplanar vectors, then C×A⋅BA⋅B×C​+C⋅A×BB⋅A×C​=.........