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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Vector Algebra

Let a =- i + j +2 k , ~b = i - j -3 k , c = a × b and d = c × a ....

Let a→=−i^+j^+2k^, b→=i^−j^−3k^,c→=a→×b→\overrightarrow{\mathrm{a}}=-\hat{i}+\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\hat{i}-\hat{j}-3 \hat{k}, \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} and d→=c→×a→\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}. Then (a⃗−b⃗)⋅d⃗(\vec{a}-\vec{b}) \cdot \vec{d} is equal to :

Options

  1. A-4
  2. B4
  3. C2
  4. D-2

Correct answer

D. -2

Step-by-step solution

c⃗=a⃗×b⃗=∣i^j^k^−1121−1−3∣=(−1,−1,0)\vec{c} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ 1 & -1 & -3 \end{vmatrix} = (-1, -1, 0).
d⃗=c⃗×a⃗=∣i^j^k^−1−10−112∣=(−2,2,−2)\vec{d} = \vec{c} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 0 \\ -1 & 1 & 2 \end{vmatrix} = (-2, 2, -2).
a⃗−b⃗=(−2,2,5)\vec{a} - \vec{b} = (-2, 2, 5).
(a⃗−b⃗)⋅d⃗=(−2)(−2)+(2)(2)+(5)(−2)=4+4−10=−2(\vec{a}-\vec{b}) \cdot \vec{d} = (-2)(-2) + (2)(2) + (5)(-2) = 4 + 4 - 10 = -2.
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