Vectors: Conceptual Doubts Cleared (12)

hard • 3 min read
Tags Vectors

Question

If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are three non-coplanar vectors, prove that [a⃗+b⃗,b⃗+c⃗,c⃗+a⃗]=2[a⃗,b⃗,c⃗][\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] = 2[\vec{a}, \vec{b}, \vec{c}], where [⋅,⋅,⋅][\cdot, \cdot, \cdot] denotes the scalar triple product.

Solution — Step by Step

[a⃗+b⃗,b⃗+c⃗,c⃗+a⃗][\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] means (a⃗+b⃗)⋅[(b⃗+c⃗)×(c⃗+a⃗)](\vec{a}+\vec{b}) \cdot [(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})].

(b⃗+c⃗)×(c⃗+a⃗)=b⃗×c⃗+b⃗×a⃗+c⃗×c⃗+c⃗×a⃗(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a}) = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a}

Since c⃗×c⃗=0\vec{c} \times \vec{c} = 0 and b⃗×a⃗=−a⃗×b⃗\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}:

=b⃗×c⃗−a⃗×b⃗+c⃗×a⃗= \vec{b} \times \vec{c} - \vec{a} \times \vec{b} + \vec{c} \times \vec{a}.

(a⃗+b⃗)⋅(b⃗×c⃗−a⃗×b⃗+c⃗×a⃗)(\vec{a}+\vec{b}) \cdot (\vec{b} \times \vec{c} - \vec{a} \times \vec{b} + \vec{c} \times \vec{a}).

Distribute and use the cyclic property a⃗⋅(b⃗×c⃗)=[a⃗,b⃗,c⃗]\vec{a} \cdot (\vec{b} \times \vec{c}) = [\vec{a},\vec{b},\vec{c}] and a⃗⋅(a⃗×b⃗)=0\vec{a} \cdot (\vec{a} \times \vec{b}) = 0 (any triple product with a repeated vector is zero).

Non-zero terms: a⃗⋅(b⃗×c⃗)=[a⃗,b⃗,c⃗]\vec{a} \cdot (\vec{b} \times \vec{c}) = [\vec{a},\vec{b},\vec{c}] and b⃗⋅(c⃗×a⃗)=[b⃗,c⃗,a⃗]=[a⃗,b⃗,c⃗]\vec{b} \cdot (\vec{c} \times \vec{a}) = [\vec{b},\vec{c},\vec{a}] = [\vec{a},\vec{b},\vec{c}] (cyclic).

So total = [a⃗,b⃗,c⃗]+[a⃗,b⃗,c⃗]=2[a⃗,b⃗,c⃗][\vec{a},\vec{b},\vec{c}] + [\vec{a},\vec{b},\vec{c}] = 2[\vec{a},\vec{b},\vec{c}].

Final answer: proved.

Why This Works

The scalar triple product is a determinant. When you replace each vector by a sum, you can split the determinant by linearity in each row — this gives 23=82^3 = 8 smaller determinants. Most of them vanish because two rows become equal (the determinant of a matrix with two equal rows is zero). Only two non-zero terms survive, both equal to [a⃗,b⃗,c⃗][\vec{a},\vec{b},\vec{c}], giving the factor 2.

This is a JEE Advanced classic — the geometric interpretation is that the parallelepiped formed by a⃗+b⃗,b⃗+c⃗,c⃗+a⃗\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a} has exactly twice the volume of the parallelepiped formed by a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}.

Alternative Method

Use determinants. Write each new vector as a linear combination, get the matrix:

M=(110011101)M = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}

(coefficients of a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} in each new vector). Compute det⁡M=2\det M = 2. Then the new triple product = det⁡M\det M times the old triple product.

For any linear combination of three vectors, the new triple product = (determinant of coefficient matrix) × (old triple product). Memorise this; it generalises far beyond this one example.

Common Mistake

Trying to expand all 8 terms term by term and getting lost. Use the linearity-of-determinants shortcut and you save 80% of the algebra.

Forgetting the cyclic property of triple products. [a⃗,b⃗,c⃗]=[b⃗,c⃗,a⃗]=[c⃗,a⃗,b⃗][\vec{a},\vec{b},\vec{c}] = [\vec{b},\vec{c},\vec{a}] = [\vec{c},\vec{a},\vec{b}], but [a⃗,c⃗,b⃗]=−[a⃗,b⃗,c⃗][\vec{a},\vec{c},\vec{b}] = -[\vec{a},\vec{b},\vec{c}] (swap two vectors → flip sign).

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

Cross Product — Find a⃗ × b⃗ Using Determinant Method

medium

Dot Product of Two Vectors — Find Angle Between Them

easy

Find the area of parallelogram with adjacent sides a=2i+3j and b=i-4j

hard

If |a|=2, |b|=3 and a·b=4, find |a×b| and angle between vectors

easy

Find Magnitude of Vector 3î + 4ĵ

easy