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Condition for Coplanarity of Vectors Defined by Four Points

Q: Why can we set the determinant of coefficients directly to zero instead of taking the cross product and dot product with a vector, b vector, c vector?

Because [AB vector AC vector AD vector] = det(M) [a vector b vector c vector], where M is the matrix of coordinates. Since a vector, b vector, c vector are non-coplanar, [a vector b vector c vector] ≠ 0, so the scalar triple product vanishes if and only if det(M) = 0.

Let a vector, b vector and c vector be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be a vector - b vector + c vector, λ a vector - 3b vector + 4c vector, -a vector + 2b vector - 3c vector and 2a vector - 4b vector + 6c vector respectively. If AB vector, AC vector and AD vector are coplanar, then λ is equal to

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Correct answer

The vectors $\vec{A B}, \vec{A C}$ and $\vec{A D}$ are coplanar when the scalar triple product $[\vec{A B} \vec{A C} \vec{A D}] = 0$ , which yields $λ = 2$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Points $A, B, C, D$ have position vectors relative to an origin: $\vec{O A} = \vec{a} - \vec{b} + \vec{c}$ $\vec{O B} = λ \vec{a} - 3 \vec{b} + 4 \vec{c}$ $\vec{O C} = - \vec{a} + 2 \vec{b} - 3 \vec{c}$ $\vec{O D} = 2 \vec{a} - 4 \vec{b} + 6 \vec{c}$ where $\vec{a}, \vec{b}, \vec{c}$ are non-zero, non-coplanar vectors, and $\vec{A B}, \vec{A C}, \vec{A D}$ are coplanar.

02Goal

Find the value of scalar $λ$ for which $\vec{A B}, \vec{A C}, \vec{A D}$ are coplanar.

03Approach

First compute the displacement vectors $\vec{A B} = \vec{O B} - \vec{O A}$ , $\vec{A C} = \vec{O C} - \vec{O A}$ , and $\vec{A D} = \vec{O D} - \vec{O A}$ as linear combinations of the basis ${\vec{a}, \vec{b}, \vec{c}}$ . Since the basis vectors are non-coplanar ( $[\vec{a} \vec{b} \vec{c}] \neq 0$ ), the coplanarity condition $[\vec{A B} \vec{A C} \vec{A D}] = 0$ is equivalent to setting the determinant of their coordinate coefficients to zero.

04Execute

Compute the displacement vectors: $\vec{A B} = (λ - 1) \vec{a} - 2 \vec{b} + 3 \vec{c}$ $\vec{A C} = - 2 \vec{a} + 3 \vec{b} - 4 \vec{c}$ $\vec{A D} = \vec{a} - 3 \vec{b} + 5 \vec{c}$

05Execute

Evaluate the scalar triple product determinant and set to zero: $∣ \begin{array}{ccc} λ - 1 & - 2 & 3 \\ - 2 & 3 & - 4 \\ 1 & - 3 & 5 \end{array} ∣ = 0$ Expanding along the first row: $(λ - 1) (15 - 12) - (- 2) (- 10 - (- 4)) + 3 (6 - 3) = 0$ $3 (λ - 1) + 2 (- 6) + 3 (3) = 0$ $3 λ - 3 - 12 + 9 = 0$ $3 λ - 6 = 0 ⟹ λ = 2$ .

✓Verify

Substitute $λ = 2$ : the first row becomes $(\begin{array}{ccc} 1 & - 2 & 3 \end{array})$ . Notice that Row 1 + Row 2 = $(\begin{array}{ccc} 1 - 2 & - 2 + 3 & 3 - 4 \end{array}) = (\begin{array}{ccc} - 1 & 1 & - 1 \end{array})$ , and Row 3 + 2*(Row 1 + Row 2) = $(\begin{array}{ccc} 1 & - 3 & 5 \end{array}) + (\begin{array}{ccc} - 2 & 2 & - 2 \end{array}) = (\begin{array}{ccc} - 1 & - 1 & 3 \end{array})$ . Direct determinant check: $1 (3) - (- 2) (- 6) + 3 (3) = 3 - 12 + 9 = 0$ . Holds identically.

Hints that build this answer step by step

What are the components of the vectors $\vec{A B}$ , $\vec{A C}$ , and $\vec{A D}$ in terms of the basis vectors $\vec{a}, \vec{b}, \vec{c}$ ?
$\vec{A B} = (λ - 1) \vec{a} - 2 \vec{b} + 3 \vec{c}$ , $\vec{A C} = - 2 \vec{a} + 3 \vec{b} - 4 \vec{c}$ , $\vec{A D} = \vec{a} - 3 \vec{b} + 5 \vec{c}$
Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, what condition ensures that $\vec{A B}, \vec{A C},$ and $\vec{A D}$ are coplanar?
The determinant formed by the coefficients of $\vec{A B}$ , $\vec{A C}$ , and $\vec{A D}$ must equal zero.
Evaluating the determinant $∣ \begin{array}{ccc} λ - 1 & - 2 & 3 \\ - 2 & 3 & - 4 \\ 1 & - 3 & 5 \end{array} ∣ = 0$ , what is the value of $λ$ ?
$λ = 2$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

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Students also ask

Why can we set the determinant of coefficients directly to zero instead of taking the cross product and dot product with

\vec{a}, \vec{b}, \vec{c}

Because $[\vec{A B} \vec{A C} \vec{A D}] = \det (M) [\vec{a} \vec{b} \vec{c}]$ , where $M$ is the matrix of coordinates. Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, $[\vec{a} \vec{b} \vec{c}] \neq 0$ , so the scalar triple product vanishes if and only if $\det (M) = 0$ .

Answer

The vectors $\vec{A B}, \vec{A C}$ and $\vec{A D}$ are coplanar when the scalar triple product $[\vec{A B} \vec{A C} \vec{A D}] = 0$ , which yields $λ = 2$ .

Why each option works or fails

Step-by-step solution

Given: Points $A, B, C, D$ have position vectors relative to an origin: $\vec{O A} = \vec{a} - \vec{b} + \vec{c}$ $\vec{O B} = λ \vec{a} - 3 \vec{b} + 4 \vec{c}$ $\vec{O C} = - \vec{a} + 2 \vec{b} - 3 \vec{c}$ $\vec{O D} = 2 \vec{a} - 4 \vec{b} + 6 \vec{c}$ where $\vec{a}, \vec{b}, \vec{c}$ are non-zero, non-coplanar vectors, and $\vec{A B}, \vec{A C}, \vec{A D}$ are coplanar.
Goal: Find the value of scalar $λ$ for which $\vec{A B}, \vec{A C}, \vec{A D}$ are coplanar.
Approach: First compute the displacement vectors $\vec{A B} = \vec{O B} - \vec{O A}$ , $\vec{A C} = \vec{O C} - \vec{O A}$ , and $\vec{A D} = \vec{O D} - \vec{O A}$ as linear combinations of the basis ${\vec{a}, \vec{b}, \vec{c}}$ . Since the basis vectors are non-coplanar ( $[\vec{a} \vec{b} \vec{c}] \neq 0$ ), the coplanarity condition $[\vec{A B} \vec{A C} \vec{A D}] = 0$ is equivalent to setting the determinant of their coordinate coefficients to zero.
Execute: Compute the displacement vectors: $\vec{A B} = (λ - 1) \vec{a} - 2 \vec{b} + 3 \vec{c}$ $\vec{A C} = - 2 \vec{a} + 3 \vec{b} - 4 \vec{c}$ $\vec{A D} = \vec{a} - 3 \vec{b} + 5 \vec{c}$
Execute: Evaluate the scalar triple product determinant and set to zero: $∣ \begin{array}{ccc} λ - 1 & - 2 & 3 \\ - 2 & 3 & - 4 \\ 1 & - 3 & 5 \end{array} ∣ = 0$ Expanding along the first row: $(λ - 1) (15 - 12) - (- 2) (- 10 - (- 4)) + 3 (6 - 3) = 0$ $3 (λ - 1) + 2 (- 6) + 3 (3) = 0$ $3 λ - 3 - 12 + 9 = 0$ $3 λ - 6 = 0 ⟹ λ = 2$ .
Verify: Substitute $λ = 2$ : the first row becomes $(\begin{array}{ccc} 1 & - 2 & 3 \end{array})$ . Notice that Row 1 + Row 2 = $(\begin{array}{ccc} 1 - 2 & - 2 + 3 & 3 - 4 \end{array}) = (\begin{array}{ccc} - 1 & 1 & - 1 \end{array})$ , and Row 3 + 2*(Row 1 + Row 2) = $(\begin{array}{ccc} 1 & - 3 & 5 \end{array}) + (\begin{array}{ccc} - 2 & 2 & - 2 \end{array}) = (\begin{array}{ccc} - 1 & - 1 & 3 \end{array})$ . Direct determinant check: $1 (3) - (- 2) (- 6) + 3 (3) = 3 - 12 + 9 = 0$ . Holds identically.

Condition for Coplanarity of Vectors Defined by Four Points

What feels right?

The vectors $\vec{A B}, \vec{A C}$ and $\vec{A D}$ are coplanar when the scalar triple product $[\vec{A B} \vec{A C} \vec{A D}] = 0$ , which yields $λ = 2$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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Condition for Coplanarity of Vectors Defined by Four Points

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The vectors $\vec{A B}, \vec{A C}$ and $\vec{A D}$ are coplanar when the scalar triple product $[\vec{A B} \vec{A C} \vec{A D}] = 0$ , which yields $λ = 2$ .

The vector $\vec{a} = - \hat{i} + 2 \hat{j} + \hat{k}$ is rotated through a right angle, passing…

For certain chemical reaction $X \to Y$ , the rate of formation of product is plot…