If [a⃗ b⃗ c⃗] = 0, the vectors are coplanar. Worked example.

Prove Vectors are Coplanar Using Scalar Triple Product

Question

Three vectors are given:

\vec{a} = \hat{i} + 2\hat{j} - \hat{k}, \quad \vec{b} = 3\hat{i} + \hat{j} + 2\hat{k}, \quad \vec{c} = 4\hat{i} - \hat{j} + \lambda\hat{k}

Find the value of $\lambda$ for which $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are coplanar. Then verify the result.

Solution — Step by Step

Three vectors are coplanar if and only if their scalar triple product equals zero:

[\vec{a} \; \vec{b} \; \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = 0

The geometric reason: if the scalar triple product is zero, the parallelepiped formed by the three vectors has zero volume — meaning all three lie in the same plane.

The scalar triple product is computed as:

[\vec{a} \; \vec{b} \; \vec{c}] = \begin{vmatrix} 1 & 2 & -1 \\ 3 & 1 & 2 \\ 4 & -1 & \lambda \end{vmatrix}

Each row is one vector’s components. Row 1 is $\vec{a}$ , row 2 is $\vec{b}$ , row 3 is $\vec{c}$ .

Expanding along the first row:

= 1 \cdot \begin{vmatrix} 1 & 2 \\ -1 & \lambda \end{vmatrix} - 2 \cdot \begin{vmatrix} 3 & 2 \\ 4 & \lambda \end{vmatrix} + (-1) \cdot \begin{vmatrix} 3 & 1 \\ 4 & -1 \end{vmatrix}

= 1(\lambda + 2) - 2(3\lambda - 8) - 1(-3 - 4)

= \lambda + 2 - 6\lambda + 16 + 7

= -5\lambda + 25

Set the scalar triple product equal to zero:

-5\lambda + 25 = 0 \implies \lambda = 5

The vectors are coplanar when $\lambda = 5$ .

Substitute $\lambda = 5$ back and recompute:

\begin{vmatrix} 1 & 2 & -1 \\ 3 & 1 & 2 \\ 4 & -1 & 5 \end{vmatrix} = (5+2) - 2(15-8) + (-1)(-3-4) = 7 - 14 + 7 = 0 \checkmark

Why This Works

The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ gives the signed volume of the parallelepiped with those three vectors as edges. When this volume is zero, the three vectors cannot span three dimensions — they’re all stuck in a plane.

Think of it physically: if you tried to build a box with these three edges and it collapsed flat, that’s coplanarity. The determinant captures exactly this — a zero determinant means the rows are linearly dependent, i.e., one vector is a linear combination of the other two.

This is one of the most direct conditions in JEE vectors. No need to find the actual plane equation unless the question explicitly asks.

Alternative Method

Instead of the determinant directly, we can use the condition that $\vec{c}$ can be expressed as $\vec{c} = x\vec{a} + y\vec{b}$ :

4\hat{i} - \hat{j} + \lambda\hat{k} = x(\hat{i} + 2\hat{j} - \hat{k}) + y(3\hat{i} + \hat{j} + 2\hat{k})

Comparing components:

$\hat{i}$ : $x + 3y = 4$
$\hat{j}$ : $2x + y = -1$
$\hat{k}$ : $-x + 2y = \lambda$

From the first two equations: multiply the second by 3 and subtract the first — $5x = -7$ , so $x = -7/5$ . Then $y = 4 - 3(-7/5) \cdot ...$

The determinant method is faster in exams. The linear combination method works but takes 2× longer. In JEE Main where time is everything, go straight for the $3 \times 3$ determinant.

Solving gives $x = 1$ , $y = 1$ (you can verify), and $\lambda = -x + 2y = -1 + 2(3) = 5$ . Same answer.

Common Mistake

Row order doesn’t matter for zero — but matters for sign. Many students panic when they get $[\vec{a}\;\vec{b}\;\vec{c}] = -[\vec{b}\;\vec{a}\;\vec{c}]$ and think they made an error. For the coplanarity condition (equals zero), swapping any two rows just changes the sign — zero stays zero. But if the question asks you to compute the scalar triple product as a value, row order matters. Here, since we’re only testing if it equals zero, any row ordering gives the correct $\lambda$ .

The second common trap: students expand the determinant correctly but forget the alternating sign pattern — the middle term gets $-2(\ldots)$ , not $+2(\ldots)$ . Always write out the cofactor signs as $+, -, +$ before expanding.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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