Let a = 7 i + j - k and b = j + 2 k . If r is a vector such that r × a + a × b = 0 and r · a = 0 , then |3 r |^2 is equal to:

Correct answer: A. 44 Given r × a + a × b = 0 r × a - b × a = 0 ( r - b ) × a = 0 r - b = λ a r = b + λ a Taking dot product with a on both sides: r · a = b · a + λ | a |^2 Since r · a = 0 , we get: λ = - b · a | a |^2 We have a = 7 i + j - k and b = j + 2 k . b · a = (0)( 7 ) + (1)(1) + (2)(-1) = -1 | a |^2 = ( 7 )^2 + (1)^2 + (-1)^2 = 9 | b |^2 = (1)^2 + (2)^2 = 5 Substituting these values: λ = -(-1)/(9) = (1)/(9) Therefore, r = b + (1)/(9) a Squaring both sides: | r |^2 = | b |^2 + (1)/(81)| a |^2 + (2)/(9)( b · a ) | r |^2 = 5 + (1)/(81)(9) + (2)/(9)(-1) | r |^2 = 5 + (1)/(9) - (2)/(9) = (44)/(9) Finally, |3 r |^2 = 9| r |^2 = 9 × (44)/(9) = 44 Answer: 44

JEE Main 20265 AprilMorning Shift

Exam: JEE Main
Year: 2026
Date: 5 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2026 · 5 April · Morning Shift · Mathematics · Vector Algebra

Let a = 7 i + j - k and b = j + 2 k . If r is a vector such that r...

Let

\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}

and

\vec{b} = \hat{j} + 2\hat{k}

. If

\vec{r}

is a vector such that

\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}

and

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

, then

|3\vec{r}|^2

is equal to:

Options

A $44$
B $54$
C $86$
D $132$

Correct answer

A. $44$

Step-by-step solution

Given

\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}

\Rightarrow \vec{r} \times \vec{a} - \vec{b} \times \vec{a} = \vec{0}

\Rightarrow (\vec{r} - \vec{b}) \times \vec{a} = \vec{0}

\Rightarrow \vec{r} - \vec{b} = \lambda \vec{a}

\Rightarrow \vec{r} = \vec{b} + \lambda \vec{a}

Taking dot product with

\vec{a}

on both sides:

\vec{r} \cdot \vec{a} = \vec{b} \cdot \vec{a} + \lambda |\vec{a}|^2

Since

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

, we get:

\lambda = -\dfrac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}

We have

\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}

and

\vec{b} = \hat{j} + 2\hat{k}

\vec{b} \cdot \vec{a} = (0)(\sqrt{7}) + (1)(1) + (2)(-1) = -1

|\vec{a}|^2 = (\sqrt{7})^2 + (1)^2 + (-1)^2 = 9

|\vec{b}|^2 = (1)^2 + (2)^2 = 5

Substituting these values:

\lambda = -\dfrac{-1}{9} = \dfrac{1}{9}

Therefore,

\vec{r} = \vec{b} + \dfrac{1}{9}\vec{a}

Squaring both sides:

|\vec{r}|^2 = |\vec{b}|^2 + \dfrac{1}{81}|\vec{a}|^2 + \dfrac{2}{9}(\vec{b} \cdot \vec{a})

|\vec{r}|^2 = 5 + \dfrac{1}{81}(9) + \dfrac{2}{9}(-1)

|\vec{r}|^2 = 5 + \dfrac{1}{9} - \dfrac{2}{9} = \dfrac{44}{9}

Finally,

|3\vec{r}|^2 = 9|\vec{r}|^2 = 9 \times \dfrac{44}{9} = 44

Answer:

44

Practice Vector Algebra on Quantrex Academy →

More from Vector Algebra