Four Coplanar Points With Position Vectors a,b,c,d, Which Identity Holds

JEE Main. Previous Year Question.

Problem

If four distinct points with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar, then which of the following is true?

1. $[\vec{a}\ \vec{b}\ \vec{c}] = [\vec{d}\ \vec{c}\ \vec{a}] + [\vec{b}\ \vec{d}\ \vec{a}] + [\vec{c}\ \vec{d}\ \vec{b}]$
2. $[\vec{a}\ \vec{b}\ \vec{c}] = [\vec{d}\ \vec{b}\ \vec{a}] + [\vec{a}\ \vec{c}\ \vec{d}] + [\vec{d}\ \vec{b}\ \vec{c}]$
3. $[\vec{a}\ \vec{b}\ \vec{c}] = [\vec{a}\ \vec{d}\ \vec{b}] + [\vec{d}\ \vec{c}\ \vec{a}] + [\vec{d}\ \vec{b}\ \vec{c}]$
4. $[\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{d}] + [\vec{d}\ \vec{a}\ \vec{c}] + [\vec{d}\ \vec{b}\ \vec{a}]$

Key insight. “Four points are coplanar” translates directly into a scalar triple product equal to zero — but only once it’s built from vectors that all originate at a single common point. Picking $d$ as that common point (rather than $a$) leads straight to an identity purely in $[\vec{a}\ \vec{b}\ \vec{c}]$.

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Approach

Since $A, B, C, D$ are coplanar, the three vectors from $D$ to the other points — $\vec{a}-\vec{d}$, $\vec{b}-\vec{d}$, $\vec{c}-\vec{d}$ — all lie in that same plane, so their scalar triple product is zero. Expanding this using the multilinearity of the scalar triple product, and discarding every term with a repeated vector (which vanishes automatically), leaves an identity relating $[\vec{a}\ \vec{b}\ \vec{c}]$ to triple products involving $\vec{d}$.

Solution

Step 1 — Set up the coplanarity condition from $D$

Since $\vec{a}-\vec{d}$, $\vec{b}-\vec{d}$, $\vec{c}-\vec{d}$ are coplanar:

$$[\vec{a}-\vec{d}\ \ \vec{b}-\vec{d}\ \ \vec{c}-\vec{d}] = 0$$

Step 2 — Expand using multilinearity

Expanding the triple product term by term:

$$[\vec{a}, \vec{b}, \vec{c}] – [\vec{a}, \vec{b}, \vec{d}] – [\vec{a}, \vec{d}, \vec{c}] + [\vec{a}, \vec{d}, \vec{d}] – [\vec{d}, \vec{b}, \vec{c}] + [\vec{d}, \vec{b}, \vec{d}] + [\vec{d}, \vec{d}, \vec{c}] – [\vec{d}, \vec{d}, \vec{d}] = 0$$

Step 3 — Drop every term with a repeated vector

Any scalar triple product with two equal vectors is automatically zero (the three vectors can’t span a parallelepiped of nonzero volume). So $[\vec{a},\vec{d},\vec{d}] = [\vec{d},\vec{b},\vec{d}] = [\vec{d},\vec{d},\vec{c}] = [\vec{d},\vec{d},\vec{d}] = 0$, leaving:

$$[\vec{a}, \vec{b}, \vec{c}] – [\vec{a}, \vec{b}, \vec{d}] – [\vec{a}, \vec{d}, \vec{c}] – [\vec{d}, \vec{b}, \vec{c}] = 0$$

$$\implies [\vec{a}, \vec{b}, \vec{c}] = [\vec{a}, \vec{b}, \vec{d}] + [\vec{a}, \vec{d}, \vec{c}] + [\vec{d}, \vec{b}, \vec{c}]$$

Step 4 — Rewrite each term using cyclic symmetry

A scalar triple product is unchanged under a cyclic permutation of its three vectors: $[\vec{u},\vec{v},\vec{w}] = [\vec{v},\vec{w},\vec{u}] = [\vec{w},\vec{u},\vec{v}]$. Applying this to each term:

$$[\vec{a},\vec{b},\vec{d}] = [\vec{b},\vec{d},\vec{a}], \qquad [\vec{a},\vec{d},\vec{c}] = [\vec{d},\vec{c},\vec{a}], \qquad [\vec{d},\vec{b},\vec{c}] = [\vec{c},\vec{d},\vec{b}]$$

Step 5 — Substitute back

$$[\vec{a}, \vec{b}, \vec{c}] = [\vec{d}, \vec{c}, \vec{a}] + [\vec{b}, \vec{d}, \vec{a}] + [\vec{c}, \vec{d}, \vec{b}]$$

This matches option 1 exactly.

Answer

Common mistakes

• Building the coplanarity condition from the wrong common point. Starting from vectors relative to $\vec{a}$ instead of $\vec{d}$ produces a different, equally valid identity — but one that won’t match this particular set of answer choices without further rearranging.
• Getting the sign wrong on a non-cyclic swap. Swapping just two vectors in a scalar triple product flips its sign; only a full cyclic rotation of all three leaves it unchanged. Mixing the two up is the easiest way to end up with the wrong option.

Practise next

• Starting from the coplanarity condition built at point $A$ instead of $D$, derive the resulting identity for $[\vec{b}\ \vec{c}\ \vec{d}]$ and check it against option 4.