Let a =- i +2 j +2 k , ~b =8 i +7 j -3 k and c be vector such that a × c = b . If c ·( i + j + k )=4 , then | a + c |^ 2 is equal to :

Correct answer: A. 27 Let c = x i + y j + z k . From a × c = b : z - y = 4 , z + 2x = 7 , y + 2x = 3 . From c · ( i + j + k ) = 4 : x + y + z = 4 . Solving: y = -1 , z = 3 , x = 2 . c = 2 i - j + 3 k . a + c = (-1+2) i + (2-1) j + (2+3) k = i + j + 5 k . | a + c |^2 = 1 + 1 + 25 = 27 .

JEE Main 202621 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Vector Algebra

Let a =- i +2 j +2 k , ~b =8 i +7 j -3 k and c be vector such that...

Let

\overrightarrow{\mathrm{a}}=-\hat{i}+2 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=8 \hat{i}+7 \hat{j}-3 \hat{k}

and

\overrightarrow{\mathrm{c}}

be vector such that

\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}

. If

\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4

, then

|\vec{a}+\vec{c}|^{2}

is equal to :

Options

Correct answer

A. 27

Step-by-step solution

Let

\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}

.
From

\vec{a} \times \vec{c} = \vec{b}

z - y = 4

z + 2x = 7

y + 2x = 3

.
From

\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4

x + y + z = 4

.
Solving:

y = -1

z = 3

x = 2

\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}

\vec{a} + \vec{c} = (-1+2)\hat{i} + (2-1)\hat{j} + (2+3)\hat{k} = \hat{i} + \hat{j} + 5\hat{k}

|\vec{a} + \vec{c}|^2 = 1 + 1 + 25 = 27

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