JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding an Unknown Vector from Vector Cross and Dot Product Relations

Let: a vector = i hat + 2j hat + 3k hat, b vector = i hat - j hat + 2k hat and c vector = 5i hat - 3j hat + 3k hat be there vectors. If r vector is a vector such that, r vector × b vector = c vector × b vector and r vector · a vector = 0, then 25|r vector|^2 is equal to

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A560B449C339D336

Step-by-step solutionView

Correct answer

Using $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ , we express $\vec{r} = \vec{c} + λ \vec{b}$ , determine $λ = - 2$ via $\vec{r} \cdot \vec{a} = 0$ , and find $25 ∣ \vec{r} ∣^{2} = 339$ .

Option analysis

Why each option works or fails

A · 560

Taking $\vec{r} = \vec{c} - λ \vec{b}$ and incorrectly solving $λ$ as $+ 2$ , leading to $\vec{r} = 5 \hat{i} - 3 \hat{j} + 3 \hat{k} + 2 (\hat{i} - \hat{j} + 2 \hat{k}) = 7 \hat{i} - 5 \hat{j} + 7 \hat{k}$ and $∣ \vec{r} ∣^{2} = 123$ . Consistently apply the condition $(\vec{c} + λ \vec{b}) \cdot \vec{a} = 0$ to get $λ = - \frac{\vec{c} \cdot \vec{a}}{\vec{b} \cdot \vec{a}} = - \frac{8}{5} \neq - 2$ .

B · 449

Computing $\vec{b} \cdot \vec{a}$ with a sign error as $1 + 2 + 6 = 9$ or $\vec{c} \cdot \vec{a}$ incorrectly, giving an incorrect scalar multiplier. Carefully calculate dot products component by component: $\vec{b} \cdot \vec{a} = (1) (1) + (- 1) (2) + (2) (3) = 5$ .

C · 339

The solution correctly establishes $\vec{r} = \vec{c} + λ \vec{b}$ , evaluates $λ = - 2$ , and calculates $25 ∣ \vec{r} ∣^{2} = 339$ . This is the correct option.

D · 336

Arithmetic slip when summing the squares of the components: calculating $9 + 1 + 1$ incorrectly or miscomputing $25 \times \frac{339}{25}$ as $336$ . Double-check the sum of squares: $3^{2} + (- 1)^{2} + (- 1)^{2} = 11$ , or if using $λ = - 8 / 5$ , ensure $(17^{2} + (- 7)^{2} + (- 1)^{2}) = 289 + 49 + 1 = 339$ .

Reviewed route

Solution

StepWorking

01Given

Given vectors $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $\vec{b} = \hat{i} - \hat{j} + 2 \hat{k}$ , and $\vec{c} = 5 \hat{i} - 3 \hat{j} + 3 \hat{k}$ . Vector $\vec{r}$ satisfies $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$ .

02Goal

Find the value of $25 ∣ \vec{r} ∣^{2}$ .

03Approach

Rewrite the cross product equation as $(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$ , which implies $\vec{r} - \vec{c}$ is collinear with $\vec{b}$ , so $\vec{r} = \vec{c} + λ \vec{b}$ . Then use the orthogonality condition $\vec{r} \cdot \vec{a} = 0$ to solve for the scalar $λ$ .

04Execute

Express $\vec{r}$ in terms of $λ$ : $\vec{r} = (5 + λ) \hat{i} - (3 + λ) \hat{j} + (3 + 2 λ) \hat{k}$ . Apply $\vec{r} \cdot \vec{a} = 0$ : $1 (5 + λ) + 2 (- 3 - λ) + 3 (3 + 2 λ) = 0$ $(5 + λ) - (6 + 2 λ) + (9 + 6 λ) = 0$ $5 λ + 8 = 0 ⟹ λ = - \frac{8}{5}$ .

05Execute

Substitute $λ = - \frac{8}{5}$ back into $\vec{r}$ : $\vec{r} = \frac{1}{5} [(25 - 8) \hat{i} - (15 - 8) \hat{j} + (15 - 16) \hat{k}] = \frac{1}{5} (17 \hat{i} - 7 \hat{j} - \hat{k})$ . Compute $25 ∣ \vec{r} ∣^{2}$ : $25 ∣ \vec{r} ∣^{2} = 17^{2} + (- 7)^{2} + (- 1)^{2} = 289 + 49 + 1 = 339$ .

✓Verify

Check dot product: $\vec{r} \cdot \vec{a} = \frac{1}{5} (17 (1) - 7 (2) - 1 (3)) = \frac{1}{5} (17 - 14 - 3) = 0$ . Matches the given condition.

Hints that build this answer step by step

Given $(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$ , how can $\vec{r}$ be represented in terms of $\vec{b}$ and $\vec{c}$ for some real scalar $λ$ ?
$\vec{r} = \vec{c} + λ \vec{b}$
Using $\vec{r} = \vec{c} + λ \vec{b}$ and the condition $\vec{r} \cdot \vec{a} = 0$ , what is the value of $λ$ ?
$λ = - \frac{8}{5}$
With $\vec{r} = \vec{c} - \frac{8}{5} \vec{b} = \frac{1}{5} (17 \hat{i} - 7 \hat{j} - \hat{k})$ , what is the value of $25 ∣ \vec{r} ∣^{2}$ ?
$339$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why does

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

imply

\vec{r} - \vec{c} = λ \vec{b}

The cross product of two non-zero vectors is zero if and only if they are parallel (collinear).

Answer

Using $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ , we express $\vec{r} = \vec{c} + λ \vec{b}$ , determine $λ = - 2$ via $\vec{r} \cdot \vec{a} = 0$ , and find $25 ∣ \vec{r} ∣^{2} = 339$ .

Why each option works or fails

A: 560 - Taking $\vec{r} = \vec{c} - λ \vec{b}$ and incorrectly solving $λ$ as $+ 2$ , leading to $\vec{r} = 5 \hat{i} - 3 \hat{j} + 3 \hat{k} + 2 (\hat{i} - \hat{j} + 2 \hat{k}) = 7 \hat{i} - 5 \hat{j} + 7 \hat{k}$ and $∣ \vec{r} ∣^{2} = 123$ . Consistently apply the condition $(\vec{c} + λ \vec{b}) \cdot \vec{a} = 0$ to get $λ = - \frac{\vec{c} \cdot \vec{a}}{\vec{b} \cdot \vec{a}} = - \frac{8}{5} \neq - 2$ .
B: 449 - Computing $\vec{b} \cdot \vec{a}$ with a sign error as $1 + 2 + 6 = 9$ or $\vec{c} \cdot \vec{a}$ incorrectly, giving an incorrect scalar multiplier. Carefully calculate dot products component by component: $\vec{b} \cdot \vec{a} = (1) (1) + (- 1) (2) + (2) (3) = 5$ .
C · correct: 339 - The solution correctly establishes $\vec{r} = \vec{c} + λ \vec{b}$ , evaluates $λ = - 2$ , and calculates $25 ∣ \vec{r} ∣^{2} = 339$ . This is the correct option.
D: 336 - Arithmetic slip when summing the squares of the components: calculating $9 + 1 + 1$ incorrectly or miscomputing $25 \times \frac{339}{25}$ as $336$ . Double-check the sum of squares: $3^{2} + (- 1)^{2} + (- 1)^{2} = 11$ , or if using $λ = - 8 / 5$ , ensure $(17^{2} + (- 7)^{2} + (- 1)^{2}) = 289 + 49 + 1 = 339$ .

Step-by-step solution

Given: Given vectors $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $\vec{b} = \hat{i} - \hat{j} + 2 \hat{k}$ , and $\vec{c} = 5 \hat{i} - 3 \hat{j} + 3 \hat{k}$ . Vector $\vec{r}$ satisfies $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$ .
Goal: Find the value of $25 ∣ \vec{r} ∣^{2}$ .
Approach: Rewrite the cross product equation as $(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$ , which implies $\vec{r} - \vec{c}$ is collinear with $\vec{b}$ , so $\vec{r} = \vec{c} + λ \vec{b}$ . Then use the orthogonality condition $\vec{r} \cdot \vec{a} = 0$ to solve for the scalar $λ$ .
Execute: Express $\vec{r}$ in terms of $λ$ : $\vec{r} = (5 + λ) \hat{i} - (3 + λ) \hat{j} + (3 + 2 λ) \hat{k}$ . Apply $\vec{r} \cdot \vec{a} = 0$ : $1 (5 + λ) + 2 (- 3 - λ) + 3 (3 + 2 λ) = 0$ $(5 + λ) - (6 + 2 λ) + (9 + 6 λ) = 0$ $5 λ + 8 = 0 ⟹ λ = - \frac{8}{5}$ .
Execute: Substitute $λ = - \frac{8}{5}$ back into $\vec{r}$ : $\vec{r} = \frac{1}{5} [(25 - 8) \hat{i} - (15 - 8) \hat{j} + (15 - 16) \hat{k}] = \frac{1}{5} (17 \hat{i} - 7 \hat{j} - \hat{k})$ . Compute $25 ∣ \vec{r} ∣^{2}$ : $25 ∣ \vec{r} ∣^{2} = 17^{2} + (- 7)^{2} + (- 1)^{2} = 289 + 49 + 1 = 339$ .
Verify: Check dot product: $\vec{r} \cdot \vec{a} = \frac{1}{5} (17 (1) - 7 (2) - 1 (3)) = \frac{1}{5} (17 - 14 - 3) = 0$ . Matches the given condition.

Finding an Unknown Vector from Vector Cross and Dot Product Relations

What feels right?

Using $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ , we express $\vec{r} = \vec{c} + λ \vec{b}$ , determine $λ = - 2$ via $\vec{r} \cdot \vec{a} = 0$ , and find $25 ∣ \vec{r} ∣^{2} = 339$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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