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Vector Algebra: JEE Main Mathematics Question with Solution Let :
a ⃗ = i ^ + 2 j ^ + 3 k ^ , b ⃗ = i ^ − j ^ + 2 k ^ \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \vec{b} = \hat{i} - \hat{j} + 2\hat{k} a = i ^ + 2 j ^ + 3 k ^ , b = i ^ − j ^ + 2 k ^ and
c ⃗ = 5 i ^ − 3 j ^ + 3 k ^ \vec{c} = 5\hat{i} - 3\hat{j} + 3\hat{k} c = 5 i ^ − 3 j ^ + 3 k ^ be there vectors. If
r ⃗ \vec{r} r is a vector such that,
r ⃗ × b ⃗ = c ⃗ × b ⃗ \vec{r} \times \vec{b} = \vec{c} \times \vec{b} r × b = c × b and
r ⃗ ⋅ a ⃗ = 0 \vec{r} \cdot \vec{a} = 0 r ⋅ a = 0 , then
25 ∣ r ⃗ ∣ 2 25|\vec{r}|^2 25∣ r ∣ 2 is equal to
Hint 1 of 3
Given ( r ⃗ − c ⃗ ) × b ⃗ = 0 ⃗ (\vec{r} - \vec{c}) \times \vec{b} = \vec{0} ( r − c ) × b = 0 r ⃗ \vec{r} r b ⃗ \vec{b} b c ⃗ \vec{c} c λ \lambda λ
r ⃗ = c ⃗ + λ b ⃗ \vec{r} = \vec{c} + \lambda\vec{b} r = c + λ b r ⃗ = λ ( c ⃗ × b ⃗ ) \vec{r} = \lambda(\vec{c} \times \vec{b}) r = λ ( c × b ) Step-by-step solution View Correct answer
Using r ⃗ × b ⃗ = c ⃗ × b ⃗ \vec{r} \times \vec{b} = \vec{c} \times \vec{b} r × b = c × b r ⃗ = c ⃗ + λ b ⃗ \vec{r} = \vec{c} + \lambda\vec{b} r = c + λ b λ = − 2 \lambda = -2 λ = − 2 r ⃗ ⋅ a ⃗ = 0 \vec{r} \cdot \vec{a} = 0 r ⋅ a = 0 25 ∣ r ⃗ ∣ 2 = 339 25|\vec{r}|^2 = 339 25∣ r ∣ 2 = 339 Option analysis
Why each option works or fails A · 560 560 560 Taking r ⃗ = c ⃗ − λ b ⃗ \vec{r} = \vec{c} - \lambda\vec{b} r = c − λ b λ \lambda λ + 2 +2 + 2 r ⃗ = 5 i ^ − 3 j ^ + 3 k ^ + 2 ( i ^ − j ^ + 2 k ^ ) = 7 i ^ − 5 j ^ + 7 k ^ \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} r = 5 i ^ − 3 j ^ + 3 k ^ + 2 ( i ^ − j ^ + 2 k ^ ) = 7 i ^ − 5 j ^ + 7 k ^ ∣ r ⃗ ∣ 2 = 123 |\vec{r}|^2 = 123 ∣ r ∣ 2 = 123 ( c ⃗ + λ b ⃗ ) ⋅ a ⃗ = 0 (\vec{c} + \lambda\vec{b}) \cdot \vec{a} = 0 ( c + λ b ) ⋅ a = 0 λ = − c ⃗ ⋅ a ⃗ b ⃗ ⋅ a ⃗ = − 8 5 ≠ − 2 \lambda = -\frac{\vec{c} \cdot \vec{a}}{\vec{b} \cdot \vec{a}} = -\frac{8}{5} \neq -2 λ = − b ⋅ a c ⋅ a = − 5 8 = − 2
B · 449 449 449 Computing b ⃗ ⋅ a ⃗ \vec{b} \cdot \vec{a} b ⋅ a 1 + 2 + 6 = 9 1 + 2 + 6 = 9 1 + 2 + 6 = 9 c ⃗ ⋅ a ⃗ \vec{c} \cdot \vec{a} c ⋅ a b ⃗ ⋅ a ⃗ = ( 1 ) ( 1 ) + ( − 1 ) ( 2 ) + ( 2 ) ( 3 ) = 5 \vec{b} \cdot \vec{a} = (1)(1) + (-1)(2) + (2)(3) = 5 b ⋅ a = ( 1 ) ( 1 ) + ( − 1 ) ( 2 ) + ( 2 ) ( 3 ) = 5
C · 339 339 339 None. The solution correctly establishes r ⃗ = c ⃗ + λ b ⃗ \vec{r} = \vec{c} + \lambda\vec{b} r = c + λ b λ = − 2 \lambda = -2 λ = − 2 25 ∣ r ⃗ ∣ 2 = 339 25|\vec{r}|^2 = 339 25∣ r ∣ 2 = 339
D · 336 336 336 Arithmetic slip when summing the squares of the components: calculating 9 + 1 + 1 9 + 1 + 1 9 + 1 + 1 25 × 339 25 25 \times \frac{339}{25} 25 × 25 339 336 336 336 3 2 + ( − 1 ) 2 + ( − 1 ) 2 = 11 3^2 + (-1)^2 + (-1)^2 = 11 3 2 + ( − 1 ) 2 + ( − 1 ) 2 = 11 λ = − 8 / 5 \lambda = -8/5 λ = − 8/5 ( 17 2 + ( − 7 ) 2 + ( − 1 ) 2 ) = 289 + 49 + 1 = 339 (17^2 + (-7)^2 + (-1)^2) = 289 + 49 + 1 = 339 ( 1 7 2 + ( − 7 ) 2 + ( − 1 ) 2 ) = 289 + 49 + 1 = 339
Step Working
01 given Given vectors a ⃗ = i ^ + 2 j ^ + 3 k ^ \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} a = i ^ + 2 j ^ + 3 k ^ b ⃗ = i ^ − j ^ + 2 k ^ \vec{b} = \hat{i} - \hat{j} + 2\hat{k} b = i ^ − j ^ + 2 k ^ c ⃗ = 5 i ^ − 3 j ^ + 3 k ^ \vec{c} = 5\hat{i} - 3\hat{j} + 3\hat{k} c = 5 i ^ − 3 j ^ + 3 k ^ r ⃗ \vec{r} r r ⃗ × b ⃗ = c ⃗ × b ⃗ \vec{r} \times \vec{b} = \vec{c} \times \vec{b} r × b = c × b r ⃗ ⋅ a ⃗ = 0 \vec{r} \cdot \vec{a} = 0 r ⋅ a = 0
02 goal Find the value of 25 ∣ r ⃗ ∣ 2 25|\vec{r}|^2 25∣ r ∣ 2
03 approach Rewrite the cross product equation as ( r ⃗ − c ⃗ ) × b ⃗ = 0 ⃗ (\vec{r} - \vec{c}) \times \vec{b} = \vec{0} ( r − c ) × b = 0 r ⃗ − c ⃗ \vec{r} - \vec{c} r − c b ⃗ \vec{b} b r ⃗ = c ⃗ + λ b ⃗ \vec{r} = \vec{c} + \lambda\vec{b} r = c + λ b r ⃗ ⋅ a ⃗ = 0 \vec{r} \cdot \vec{a} = 0 r ⋅ a = 0 λ \lambda λ
04 execute Express r ⃗ \vec{r} r λ \lambda λ r ⃗ = ( 5 + λ ) i ^ − ( 3 + λ ) j ^ + ( 3 + 2 λ ) k ^ \vec{r} = (5 + \lambda)\hat{i} - (3 + \lambda)\hat{j} + (3 + 2\lambda)\hat{k} r = ( 5 + λ ) i ^ − ( 3 + λ ) j ^ + ( 3 + 2 λ ) k ^ r ⃗ ⋅ a ⃗ = 0 \vec{r} \cdot \vec{a} = 0 r ⋅ a = 0 1 ( 5 + λ ) + 2 ( − 3 − λ ) + 3 ( 3 + 2 λ ) = 0 1(5 + \lambda) + 2(-3 - \lambda) + 3(3 + 2\lambda) = 0 1 ( 5 + λ ) + 2 ( − 3 − λ ) + 3 ( 3 + 2 λ ) = 0 ( 5 + λ ) − ( 6 + 2 λ ) + ( 9 + 6 λ ) = 0 (5 + \lambda) - (6 + 2\lambda) + (9 + 6\lambda) = 0 ( 5 + λ ) − ( 6 + 2 λ ) + ( 9 + 6 λ ) = 0 5 λ + 8 = 0 ⟹ λ = − 8 5 5\lambda + 8 = 0 \implies \lambda = -\frac{8}{5} 5 λ + 8 = 0 ⟹ λ = − 5 8
05 execute Substitute λ = − 8 5 \lambda = -\frac{8}{5} λ = − 5 8 r ⃗ \vec{r} r r ⃗ = 1 5 [ ( 25 − 8 ) i ^ − ( 15 − 8 ) j ^ + ( 15 − 16 ) k ^ ] = 1 5 ( 17 i ^ − 7 j ^ − k ^ ) \vec{r} = \frac{1}{5}\left[(25 - 8)\hat{i} - (15 - 8)\hat{j} + (15 - 16)\hat{k}\right] = \frac{1}{5}(17\hat{i} - 7\hat{j} - \hat{k}) r = 5 1 [ ( 25 − 8 ) i ^ − ( 15 − 8 ) j ^ + ( 15 − 16 ) k ^ ] = 5 1 ( 17 i ^ − 7 j ^ − k ^ ) 25 ∣ r ⃗ ∣ 2 25|\vec{r}|^2 25∣ r ∣ 2 25 ∣ r ⃗ ∣ 2 = 17 2 + ( − 7 ) 2 + ( − 1 ) 2 = 289 + 49 + 1 = 339 25|\vec{r}|^2 = 17^2 + (-7)^2 + (-1)^2 = 289 + 49 + 1 = 339 25∣ r ∣ 2 = 1 7 2 + ( − 7 ) 2 + ( − 1 ) 2 = 289 + 49 + 1 = 339
✓ verify Check dot product: r ⃗ ⋅ a ⃗ = 1 5 ( 17 ( 1 ) − 7 ( 2 ) − 1 ( 3 ) ) = 1 5 ( 17 − 14 − 3 ) = 0 \vec{r} \cdot \vec{a} = \frac{1}{5}(17(1) - 7(2) - 1(3)) = \frac{1}{5}(17 - 14 - 3) = 0 r ⋅ a = 5 1 ( 17 ( 1 ) − 7 ( 2 ) − 1 ( 3 )) = 5 1 ( 17 − 14 − 3 ) = 0
Hints that build this answer step by step Given ( r ⃗ − c ⃗ ) × b ⃗ = 0 ⃗ (\vec{r} - \vec{c}) \times \vec{b} = \vec{0} ( r − c ) × b = 0 r ⃗ \vec{r} r b ⃗ \vec{b} b c ⃗ \vec{c} c λ \lambda λ
r ⃗ = c ⃗ + λ b ⃗ \vec{r} = \vec{c} + \lambda\vec{b} r = c + λ b Using r ⃗ = c ⃗ + λ b ⃗ \vec{r} = \vec{c} + \lambda\vec{b} r = c + λ b r ⃗ ⋅ a ⃗ = 0 \vec{r} \cdot \vec{a} = 0 r ⋅ a = 0 λ \lambda λ
λ = − 8 5 \lambda = -\frac{8}{5} λ = − 5 8 With r ⃗ = c ⃗ − 8 5 b ⃗ = 1 5 ( 17 i ^ − 7 j ^ − k ^ ) \vec{r} = \vec{c} - \frac{8}{5}\vec{b} = \frac{1}{5}(17\hat{i} - 7\hat{j} - \hat{k}) r = c − 5 8 b = 5 1 ( 17 i ^ − 7 j ^ − k ^ ) 25 ∣ r ⃗ ∣ 2 25|\vec{r}|^2 25∣ r ∣ 2
339 339 339 Your next move We think you should solve this next Same idea, new question If the four points, whose position vectors are… Solve → Same idea, new question Let a ⃗ \vec{\mathrm{a}} a b ⃗ \vec{\mathrm{b}} b Solve → Same idea, new question If a ⃗ , b ⃗ , c ⃗ \vec{a}, \vec{b}, \vec{c} a , b , c n ^ \hat{n} n ^ Solve → ✓ Source and academic review↓
Question type Single correct
Exam relevance JEE Main · Mathematics
Concepts assessed Mathematics
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Why does ( r ⃗ − c ⃗ ) × b ⃗ = 0 ⃗ (\vec{r} - \vec{c}) \times \vec{b} = \vec{0} ( r − c ) × b = 0 r ⃗ − c ⃗ = λ b ⃗ \vec{r} - \vec{c} = \lambda\vec{b} r − c = λ b The cross product of two non-zero vectors is zero if and only if they are parallel (collinear).
Answer Using r ⃗ × b ⃗ = c ⃗ × b ⃗ \vec{r} \times \vec{b} = \vec{c} \times \vec{b} r × b = c × b r ⃗ = c ⃗ + λ b ⃗ \vec{r} = \vec{c} + \lambda\vec{b} r = c + λ b λ = − 2 \lambda = -2 λ = − 2 r ⃗ ⋅ a ⃗ = 0 \vec{r} \cdot \vec{a} = 0 r ⋅ a = 0 25 ∣ r ⃗ ∣ 2 = 339 25|\vec{r}|^2 = 339 25∣ r ∣ 2 = 339
Why each option works or fails A: 560 560 560 - Taking r ⃗ = c ⃗ − λ b ⃗ \vec{r} = \vec{c} - \lambda\vec{b} r = c − λ b λ \lambda λ + 2 +2 + 2 r ⃗ = 5 i ^ − 3 j ^ + 3 k ^ + 2 ( i ^ − j ^ + 2 k ^ ) = 7 i ^ − 5 j ^ + 7 k ^ \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} r = 5 i ^ − 3 j ^ + 3 k ^ + 2 ( i ^ − j ^ + 2 k ^ ) = 7 i ^ − 5 j ^ + 7 k ^ ∣ r ⃗ ∣ 2 = 123 |\vec{r}|^2 = 123 ∣ r ∣ 2 = 123 ( c ⃗ + λ b ⃗ ) ⋅ a ⃗ = 0 (\vec{c} + \lambda\vec{b}) \cdot \vec{a} = 0 ( c + λ b ) ⋅ a = 0 λ = − c ⃗ ⋅ a ⃗ b ⃗ ⋅ a ⃗ = − 8 5 ≠ − 2 \lambda = -\frac{\vec{c} \cdot \vec{a}}{\vec{b} \cdot \vec{a}} = -\frac{8}{5} \neq -2 λ = − b ⋅ a c ⋅ a = − 5 8 = − 2 B: 449 449 449 - Computing b ⃗ ⋅ a ⃗ \vec{b} \cdot \vec{a} b ⋅ a 1 + 2 + 6 = 9 1 + 2 + 6 = 9 1 + 2 + 6 = 9 c ⃗ ⋅ a ⃗ \vec{c} \cdot \vec{a} c ⋅ a b ⃗ ⋅ a ⃗ = ( 1 ) ( 1 ) + ( − 1 ) ( 2 ) + ( 2 ) ( 3 ) = 5 \vec{b} \cdot \vec{a} = (1)(1) + (-1)(2) + (2)(3) = 5 b ⋅ a = ( 1 ) ( 1 ) + ( − 1 ) ( 2 ) + ( 2 ) ( 3 ) = 5 C · correct: 339 339 339 - None. The solution correctly establishes r ⃗ = c ⃗ + λ b ⃗ \vec{r} = \vec{c} + \lambda\vec{b} r = c + λ b λ = − 2 \lambda = -2 λ = − 2 25 ∣ r ⃗ ∣ 2 = 339 25|\vec{r}|^2 = 339 25∣ r ∣ 2 = 339 D: 336 336 336 - Arithmetic slip when summing the squares of the components: calculating 9 + 1 + 1 9 + 1 + 1 9 + 1 + 1 25 × 339 25 25 \times \frac{339}{25} 25 × 25 339 336 336 336 3 2 + ( − 1 ) 2 + ( − 1 ) 2 = 11 3^2 + (-1)^2 + (-1)^2 = 11 3 2 + ( − 1 ) 2 + ( − 1 ) 2 = 11 λ = − 8 / 5 \lambda = -8/5 λ = − 8/5 ( 17 2 + ( − 7 ) 2 + ( − 1 ) 2 ) = 289 + 49 + 1 = 339 (17^2 + (-7)^2 + (-1)^2) = 289 + 49 + 1 = 339 ( 1 7 2 + ( − 7 ) 2 + ( − 1 ) 2 ) = 289 + 49 + 1 = 339 Step-by-step solution given: Given vectors a ⃗ = i ^ + 2 j ^ + 3 k ^ \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} a = i ^ + 2 j ^ + 3 k ^ b ⃗ = i ^ − j ^ + 2 k ^ \vec{b} = \hat{i} - \hat{j} + 2\hat{k} b = i ^ − j ^ + 2 k ^ c ⃗ = 5 i ^ − 3 j ^ + 3 k ^ \vec{c} = 5\hat{i} - 3\hat{j} + 3\hat{k} c = 5 i ^ − 3 j ^ + 3 k ^ r ⃗ \vec{r} r r ⃗ × b ⃗ = c ⃗ × b ⃗ \vec{r} \times \vec{b} = \vec{c} \times \vec{b} r × b = c × b r ⃗ ⋅ a ⃗ = 0 \vec{r} \cdot \vec{a} = 0 r ⋅ a = 0 goal: Find the value of 25 ∣ r ⃗ ∣ 2 25|\vec{r}|^2 25∣ r ∣ 2 approach: Rewrite the cross product equation as ( r ⃗ − c ⃗ ) × b ⃗ = 0 ⃗ (\vec{r} - \vec{c}) \times \vec{b} = \vec{0} ( r − c ) × b = 0 r ⃗ − c ⃗ \vec{r} - \vec{c} r − c b ⃗ \vec{b} b r ⃗ = c ⃗ + λ b ⃗ \vec{r} = \vec{c} + \lambda\vec{b} r = c + λ b r ⃗ ⋅ a ⃗ = 0 \vec{r} \cdot \vec{a} = 0 r ⋅ a = 0 λ \lambda λ execute: Express r ⃗ \vec{r} r λ \lambda λ r ⃗ = ( 5 + λ ) i ^ − ( 3 + λ ) j ^ + ( 3 + 2 λ ) k ^ \vec{r} = (5 + \lambda)\hat{i} - (3 + \lambda)\hat{j} + (3 + 2\lambda)\hat{k} r = ( 5 + λ ) i ^ − ( 3 + λ ) j ^ + ( 3 + 2 λ ) k ^ r ⃗ ⋅ a ⃗ = 0 \vec{r} \cdot \vec{a} = 0 r ⋅ a = 0 1 ( 5 + λ ) + 2 ( − 3 − λ ) + 3 ( 3 + 2 λ ) = 0 1(5 + \lambda) + 2(-3 - \lambda) + 3(3 + 2\lambda) = 0 1 ( 5 + λ ) + 2 ( − 3 − λ ) + 3 ( 3 + 2 λ ) = 0 ( 5 + λ ) − ( 6 + 2 λ ) + ( 9 + 6 λ ) = 0 (5 + \lambda) - (6 + 2\lambda) + (9 + 6\lambda) = 0 ( 5 + λ ) − ( 6 + 2 λ ) + ( 9 + 6 λ ) = 0 5 λ + 8 = 0 ⟹ λ = − 8 5 5\lambda + 8 = 0 \implies \lambda = -\frac{8}{5} 5 λ + 8 = 0 ⟹ λ = − 5 8 execute: Substitute λ = − 8 5 \lambda = -\frac{8}{5} λ = − 5 8 r ⃗ \vec{r} r r ⃗ = 1 5 [ ( 25 − 8 ) i ^ − ( 15 − 8 ) j ^ + ( 15 − 16 ) k ^ ] = 1 5 ( 17 i ^ − 7 j ^ − k ^ ) \vec{r} = \frac{1}{5}\left[(25 - 8)\hat{i} - (15 - 8)\hat{j} + (15 - 16)\hat{k}\right] = \frac{1}{5}(17\hat{i} - 7\hat{j} - \hat{k}) r = 5 1 [ ( 25 − 8 ) i ^ − ( 15 − 8 ) j ^ + ( 15 − 16 ) k ^ ] = 5 1 ( 17 i ^ − 7 j ^ − k ^ ) 25 ∣ r ⃗ ∣ 2 25|\vec{r}|^2 25∣ r ∣ 2 25 ∣ r ⃗ ∣ 2 = 17 2 + ( − 7 ) 2 + ( − 1 ) 2 = 289 + 49 + 1 = 339 25|\vec{r}|^2 = 17^2 + (-7)^2 + (-1)^2 = 289 + 49 + 1 = 339 25∣ r ∣ 2 = 1 7 2 + ( − 7 ) 2 + ( − 1 ) 2 = 289 + 49 + 1 = 339 verify: Check dot product: r ⃗ ⋅ a ⃗ = 1 5 ( 17 ( 1 ) − 7 ( 2 ) − 1 ( 3 ) ) = 1 5 ( 17 − 14 − 3 ) = 0 \vec{r} \cdot \vec{a} = \frac{1}{5}(17(1) - 7(2) - 1(3)) = \frac{1}{5}(17 - 14 - 3) = 0 r ⋅ a = 5 1 ( 17 ( 1 ) − 7 ( 2 ) − 1 ( 3 )) = 5 1 ( 17 − 14 − 3 ) = 0