If the vectors b, c, d are not...

MathematicsVector Triple ProductJEE Advanced 1994Moderate

If the vectors $b, c, d$ are not coplanar, then prove that the vector $(a \times b) \times (c \times d) + (a \times c) \times (d \times b) + (a \times d) \times (b \times c)$ is parallel to $a$ .

Visualized Solution (English)

Visualizing the Vectors

Given vectors: $a, b, c, d$
Constraint: $b, c, d$ are non-coplanar ( $[b c d] \neq = 0$ )
Objective: Prove the sum of three vector quadruple products is parallel to $a$

The Vector Quadruple Product Identity

Identity: $(x \times y) \times (z \times w) = [x z w] y - [y z w] x$
This identity allows us to expand each term of the given expression into a linear combination of the first two vectors.

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a \times b) \times (c \times d)

Applying identity to the first term:
$(a \times b) \times (c \times d) = [a c d] b - [b c d] a$

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a \times c) \times (d \times b)

Applying identity to the second term:
$(a \times c) \times (d \times b) = [a d b] c - [c d b] a$

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a \times d) \times (b \times c)

Applying identity to the third term:
$(a \times d) \times (b \times c) = [a b c] d - [d b c] a$

Summing the Results

Summing all three expanded terms:
Sum $= ([a c d] b + [a d b] c + [a b c] d) - ([b c d] + [c d b] + [d b c]) a$

Cyclic Property of STP

Cyclic property: $[b c d] = [c d b] = [d b c]$
Therefore, the second part of the sum simplifies to: $- 3 [b c d] a$

The Basis Expansion Identity

Any vector $a$ can be expressed in terms of non-coplanar vectors $b, c, d$ as:
$[b c d] a = [a c d] b + [a d b] c + [a b c] d$
This identity relates the first part of our sum directly to vector $a$ .

Final Conclusion

Substituting the identity into the sum:
Sum $= [b c d] a - 3 [b c d] a$
Sum $= - 2 [b c d] a$
Since the result is a scalar multiple of $a$ , it is parallel to $a$ .

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