Let a =2 i + j -2 k , b = i + j and c = a × b . Let d be a vector such that | d - a |= 11 ,| c × d |=3 and the angle between c and d is (π)/(4) . Then a · d is equal to

Correct answer: C. 0 c = a × b = vmatrix i & j & k 2 & 1 & -2 1 & 1 & 0 vmatrix = 2 i - 2 j + k | c | = 4+4+1 = 3 , | a | = 4+1+4 = 3 From | c × d | = 3 and ( ( c , d )) = 1 2 : | c || d | (π)/(4) = 3 → 3| d | · 1 2 = 3 → | d | = 2 From | d - a |^2 = 11 : | d |^2 - 2 a · d + | a |^2 = 11 2 - 2 a · d + 9 = 11 a · d = 0

JEE Main 202624 JanuaryMorning Shift

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

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

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

Let

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

and

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

. Let

\vec{d}

be a vector such that

|\vec{d}-\vec{a}|=\sqrt{11},|\vec{c} \times \vec{d}|=3

and the angle between

\vec{c}

and

\vec{d}

\frac{\pi}{4}

. Then

\vec{a} \cdot \vec{d}

is equal to

Options

Correct answer

C. 0

Step-by-step solution

\vec{c} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = 2\hat{i} - 2\hat{j} + \hat{k}

|\vec{c}| = \sqrt{4+4+1} = 3

|\vec{a}| = \sqrt{4+1+4} = 3

From

|\vec{c} \times \vec{d}| = 3

and

\cos(\angle(\vec{c}, \vec{d})) = \frac{1}{\sqrt{2}}

|\vec{c}||\vec{d}|\sin\frac{\pi}{4} = 3

→

3|\vec{d}| \cdot \frac{1}{\sqrt{2}} = 3

→

|\vec{d}| = \sqrt{2}

From

|\vec{d} - \vec{a}|^2 = 11

|\vec{d}|^2 - 2\vec{a} \cdot \vec{d} + |\vec{a}|^2 = 11

2 - 2\vec{a} \cdot \vec{d} + 9 = 11

\vec{a} \cdot \vec{d} = 0

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