Let a =2 i -5 j +5 k and b = i - j +3 k . If c is a vector such that 2( a × c )+3( b × c )= 0 and ( a - b ) · c =-97 , then | c × k |^ 2 is equal to

Correct answer: B. 218 2( a × c ) + 3( b × c ) = 0 (2 a + 3 b ) × c = 0 . So c (2 a + 3 b ) = (7, -13, 19) . Let c = t(7, -13, 19) . a - b = (1, -4, 2) . ( a - b ) · c = t(7 + 52 + 38) = 97t = -97 t = -1 . c = (-7, 13, -19) . c × k = vmatrix i & j & k -7 & 13 & -19 0 & 0 & 1 vmatrix = (13, 7, 0) . | c × k |^2 = 169 + 49 = 218 .

JEE Main 202624 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Vector Algebra

Let a =2 i -5 j +5 k and b = i - j +3 k . If c is a vector such...

Let

\vec{a}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}

and

\vec{b}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}

. If

\vec{c}

is a vector such that

2(\vec{a} \times \vec{c})+3(\vec{b} \times \vec{c})=\overrightarrow{0}

and

(\vec{a}-\vec{b}) \cdot \vec{c}=-97

, then

|\vec{c} \times \hat{\mathrm{k}}|^{2}

is equal to

Options

A233
B218
C193
D205

Correct answer

B. 218

Step-by-step solution

2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0} \Rightarrow (2\vec{a} + 3\vec{b}) \times \vec{c} = \vec{0}

.
So

\vec{c} \parallel (2\vec{a} + 3\vec{b}) = (7, -13, 19)

. Let

\vec{c} = t(7, -13, 19)

\vec{a} - \vec{b} = (1, -4, 2)

(\vec{a} - \vec{b}) \cdot \vec{c} = t(7 + 52 + 38) = 97t = -97 \Rightarrow t = -1

\vec{c} = (-7, 13, -19)

\vec{c} \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -7 & 13 & -19 \\ 0 & 0 & 1 \end{vmatrix} = (13, 7, 0)

|\vec{c} \times \hat{k}|^2 = 169 + 49 = 218

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