JEE MainMathematics

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Vector Algebra: JEE Main Mathematics Question with Solution

If the four points, whose position vectors are

3\hat{\mathrm{i}}-4\hat{\mathrm{j}}+2\hat{\mathrm{k}}

\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}

-2\hat{\mathrm{i}}-\hat{\mathrm{j}}+3\hat{\mathrm{k}}

and

5\hat{\mathrm{i}}-2\alpha\hat{\mathrm{j}}+4\hat{\mathrm{k}}

are coplanar, then

\alpha

is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{73}{17}

\frac{107}{17}

\frac{-73}{17}

\frac{-107}{17}

Step-by-step solutionView

Correct answer

Four points are coplanar when the scalar triple product of three vectors formed by a common reference point is zero, giving $\alpha = \frac{73}{17}$ .

Option analysis

Why each option works or fails

A ·

\frac{73}{17}

None. The student correctly applies the coplanarity condition using the scalar triple product and solves the resulting linear equation. Correctly setting the determinant of vectors $\vec{AB}$ , $\vec{AC}$ , and $\vec{AD}$ to zero yields $34\alpha - 146 = 0$ , giving $\alpha = \frac{73}{17}$ .

B ·

\frac{107}{17}

The student made a sign error when expanding the determinant, likely treating the middle column expansion or a component subtraction with the wrong sign. Double-check vector differences and determinant minors; expanding with correct signs gives $34\alpha = 146$ , not $34\alpha = 214$ .

C ·

\frac{-73}{17}

The student dropped a negative sign when isolating $\alpha$ in the linear equation $34\alpha - 146 = 0$ or inverted the sign of the constant term. When moving $-146$ to the other side of the equation $34\alpha - 146 = 0$ , it becomes $+146$ , resulting in a positive value $\alpha = \frac{73}{17}$ .

D ·

\frac{-107}{17}

The student committed both an expansion sign error and an inversion error while solving the final linear equation. Carefully compute cofactors and keep track of signs at each step of the determinant expansion.

Reviewed route

Solution

StepWorking

01given

Four points $P(3, -4, 2)$ , $Q(1, 2, -1)$ , $R(-2, -1, 3)$ , and $S(5, -2\alpha, 4)$ are coplanar.

02goal

Find the value of the parameter $\alpha$ .

03approach

Four points $P, Q, R, S$ are coplanar if and only if three vectors formed by them are coplanar. This means their scalar triple product is zero: $[\vec{PQ}, \vec{QR}, \vec{RS}] = 0$ . Equivalently, the determinant of their components vanishes.

04execute

Compute the vectors: $\vec{PQ} = (1 - 3)\hat{i} + (2 - (-4))\hat{j} + (-1 - 2)\hat{k} = -2\hat{i} + 6\hat{j} - 3\hat{k}$ $\vec{QR} = (-2 - 1)\hat{i} + (-1 - 2)\hat{j} + (3 - (-1))\hat{k} = -3\hat{i} - 3\hat{j} + 4\hat{k}$ $\vec{RS} = (5 - (-2))\hat{i} + (-2\alpha - (-1))\hat{j} + (4 - 3)\hat{k} = 7\hat{i} + (1 - 2\alpha)\hat{j} + \hat{k}$

05execute

Set the determinant equal to 0 and expand: $\begin{vmatrix} -2 & 6 & -3 \\ -3 & -3 & 4 \\ 7 & 1-2\alpha & 1 \end{vmatrix} = 0$ Expanding along the first row: $-2[(-3)(1) - 4(1 - 2\alpha)] - 6[(-3)(1) - (4)(7)] - 3[(-3)(1 - 2\alpha) - (-3)(7)] = 0$ $-2[-3 - 4 + 8\alpha] - 6[-3 - 28] - 3[-3 + 6\alpha + 21] = 0$ $-2(8\alpha - 7) - 6(-31) - 3(6\alpha + 18) = 0$ $-16\alpha + 14 + 186 - 18\alpha - 54 = 0$ $-34\alpha + 146 = 0 \implies 34\alpha = 146 \implies \alpha = \frac{146}{34} = \frac{73}{17}$

✓verify

Substitute $\alpha = 73/17$ back into the 2nd component of $\vec{RS}$ : $1 - 2(73/17) = -129/17$ . The determinant evaluates to $-34(73/17) + 146 = -146 + 146 = 0$ .

Hints that build this answer step by step

Let the points be $A(3, -4, 2)$ , $B(1, 2, -1)$ , $C(-2, -1, 3)$ , and $D(5, -2\alpha, 4)$ . Which vectors can we form from the common initial point $A$ to test for coplanarity?
$\vec{AB} = (-2, 6, -3)$ , $\vec{AC} = (-5, 3, 1)$ , and $\vec{AD} = (2, 4 - 2\alpha, 2)$
What condition must these three vectors $\vec{AB}$ , $\vec{AC}$ , and $\vec{AD}$ satisfy because the four points are coplanar?
Their scalar triple product must be zero: $\det \begin{pmatrix} -2 & 6 & -3 \\ -5 & 3 & 1 \\ 2 & 4 - 2\alpha & 2 \end{pmatrix} = 0$
Expanding the determinant $\begin{vmatrix} -2 & 6 & -3 \\ -5 & 3 & 1 \\ 2 & 4 - 2\alpha & 2 \end{vmatrix} = 0$ , what linear equation in $\alpha$ do we obtain?
$34\alpha - 146 = 0$
Solving $34\alpha - 146 = 0$ for $\alpha$ , what is the final value?
$\alpha = \frac{73}{17}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Does it matter which three vectors connecting the four points we choose?

No, any three linearly independent combinations connecting all 4 points span the same subspace. For example, you can use PQ, PR, PS or PQ, QR, RS. Both sets yield the same value for alpha.

Answer

Four points are coplanar when the scalar triple product of three vectors formed by a common reference point is zero, giving $\alpha = \frac{73}{17}$ .

Why each option works or fails

A · correct: $\frac{73}{17}$ - None. The student correctly applies the coplanarity condition using the scalar triple product and solves the resulting linear equation. Correctly setting the determinant of vectors $\vec{AB}$ , $\vec{AC}$ , and $\vec{AD}$ to zero yields $34\alpha - 146 = 0$ , giving $\alpha = \frac{73}{17}$ .
B: $\frac{107}{17}$ - The student made a sign error when expanding the determinant, likely treating the middle column expansion or a component subtraction with the wrong sign. Double-check vector differences and determinant minors; expanding with correct signs gives $34\alpha = 146$ , not $34\alpha = 214$ .
C: $\frac{-73}{17}$ - The student dropped a negative sign when isolating $\alpha$ in the linear equation $34\alpha - 146 = 0$ or inverted the sign of the constant term. When moving $-146$ to the other side of the equation $34\alpha - 146 = 0$ , it becomes $+146$ , resulting in a positive value $\alpha = \frac{73}{17}$ .
D: $\frac{-107}{17}$ - The student committed both an expansion sign error and an inversion error while solving the final linear equation. Carefully compute cofactors and keep track of signs at each step of the determinant expansion.

Step-by-step solution

given: Four points $P(3, -4, 2)$ , $Q(1, 2, -1)$ , $R(-2, -1, 3)$ , and $S(5, -2\alpha, 4)$ are coplanar.
goal: Find the value of the parameter $\alpha$ .
approach: Four points $P, Q, R, S$ are coplanar if and only if three vectors formed by them are coplanar. This means their scalar triple product is zero: $[\vec{PQ}, \vec{QR}, \vec{RS}] = 0$ . Equivalently, the determinant of their components vanishes.
execute: Compute the vectors: $\vec{PQ} = (1 - 3)\hat{i} + (2 - (-4))\hat{j} + (-1 - 2)\hat{k} = -2\hat{i} + 6\hat{j} - 3\hat{k}$ $\vec{QR} = (-2 - 1)\hat{i} + (-1 - 2)\hat{j} + (3 - (-1))\hat{k} = -3\hat{i} - 3\hat{j} + 4\hat{k}$ $\vec{RS} = (5 - (-2))\hat{i} + (-2\alpha - (-1))\hat{j} + (4 - 3)\hat{k} = 7\hat{i} + (1 - 2\alpha)\hat{j} + \hat{k}$
execute: Set the determinant equal to 0 and expand: $\begin{vmatrix} -2 & 6 & -3 \\ -3 & -3 & 4 \\ 7 & 1-2\alpha & 1 \end{vmatrix} = 0$ Expanding along the first row: $-2[(-3)(1) - 4(1 - 2\alpha)] - 6[(-3)(1) - (4)(7)] - 3[(-3)(1 - 2\alpha) - (-3)(7)] = 0$ $[\vec{PQ}, \vec{QR}, \vec{RS}] = 0$ $-2(8\alpha - 7) - 6(-31) - 3(6\alpha + 18) = 0$ $-16\alpha + 14 + 186 - 18\alpha - 54 = 0$ $-34\alpha + 146 = 0 \implies 34\alpha = 146 \implies \alpha = \frac{146}{34} = \frac{73}{17}$
verify: Substitute $\alpha = 73/17$ back into the 2nd component of $\vec{RS}$ : $1 - 2(73/17) = -129/17$ . The determinant evaluates to $-34(73/17) + 146 = -146 + 146 = 0$ .

Vector Algebra: JEE Main Mathematics Question with Solution

What feels right?

Four points are coplanar when the scalar triple product of three vectors formed by a common reference point is zero, giving $\alpha = \frac{73}{17}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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