PYQs Homework Notes Mocks Board Daily Homework MHT-CET Maths Question type Convert the Cartesian equations of a line to vector or symmetric form This type of question, asked 6 times: 2 May 2023 S1, 4 May 2023 S2, 9 May 2024 S1, 11 May 2024 S1, 16 Apr 2026 S1, 19 Apr 2026 S1. All 6 questions below, oldest first.
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Question 1 2 May 2023 S1
101 The vector equation of a line whose Cartesian equations are y = 2 , 4 x − 3 z + 5 = 0 y = 2, 4x - 3z + 5 = 0 y = 2 , 4 x − 3 z + 5 = 0 is A. r ⃗ = ( 3 i ^ + 4 k ^ ) + λ ( 2 j ^ + 5 3 k ^ ) \vec{r} = (3\hat{i} + 4\hat{k}) + \lambda(2\hat{j} + \frac{5}{3}\hat{k}) r = ( 3 i ^ + 4 k ^ ) + λ ( 2 j ^ + 3 5 k ^ ) B. r ⃗ = ( 3 i ^ + 4 k ^ ) + λ ( 2 j ^ − 5 3 k ^ ) \vec{r} = (3\hat{i} + 4\hat{k}) + \lambda(2\hat{j} - \frac{5}{3}\hat{k}) r = ( 3 i ^ + 4 k ^ ) + λ ( 2 j ^ − 3 5 k ^ ) C. r ⃗ = 2 j ^ + 5 3 k ^ + λ ( 3 i ^ + 4 k ^ ) \vec{r} = 2\hat{j} + \frac{5}{3}\hat{k} + \lambda(3\hat{i} + 4\hat{k}) r = 2 j ^ + 3 5 k ^ + λ ( 3 i ^ + 4 k ^ ) D. r ⃗ = 2 j ^ − 5 3 k ^ + λ ( 3 i ^ + 4 k ^ ) \vec{r} = 2\hat{j} - \frac{5}{3}\hat{k} + \lambda(3\hat{i} + 4\hat{k}) r = 2 j ^ − 3 5 k ^ + λ ( 3 i ^ + 4 k ^ ) Tap an option to check your answer.
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Question 2 4 May 2023 S2
121 The Cartesian equation of a line is 2 x − 2 = 3 y + 1 = 6 z − 2 2x-2=3y+1=6z-2 2 x − 2 = 3 y + 1 = 6 z − 2 , then the vector equation of the line is A. r ⃗ = ( i ^ − j ^ 3 + k ^ 3 ) + λ ( 3 i ^ + 2 j ^ + k ^ ) \vec{r} = (\hat{i}-\frac{\hat{j}}{3}+\frac{\hat{k}}{3})+\lambda(3\hat{i}+2\hat{j}+\hat{k}) r = ( i ^ − 3 j ^ + 3 k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) B. r ⃗ = ( − i ^ + j ^ 3 − k ^ 3 ) + λ ( 1 2 i ^ + 1 3 j ^ + 1 6 k ^ ) \vec{r} = (-\hat{i}+\frac{\hat{j}}{3}-\frac{\hat{k}}{3})+\lambda(\frac{1}{2}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{6}\hat{k}) r = ( − i ^ + 3 j ^ − 3 k ^ ) + λ ( 2 1 i ^ + 3 1 j ^ + 6 1 k ^ ) C. r ⃗ = ( 3 i ^ − j ^ − k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) \vec{r} = (3\hat{i}-\hat{j}-\hat{k})+\lambda(3\hat{i}+2\hat{j}+\hat{k}) r = ( 3 i ^ − j ^ − k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) D. r ⃗ = ( i ^ − j ^ + k ^ ) + λ ( 1 2 i ^ + 1 3 j ^ + 1 6 k ^ ) \vec{r} = (\hat{i}-\hat{j}+\hat{k})+\lambda(\frac{1}{2}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{6}\hat{k}) r = ( i ^ − j ^ + k ^ ) + λ ( 2 1 i ^ + 3 1 j ^ + 6 1 k ^ ) Tap an option to check your answer.
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Question 3 9 May 2024 S1
131 If the Cartesian equation of a line is 6 x − 2 = 3 y + 1 = 2 z − 2 6x-2=3y+1=2z-2 6 x − 2 = 3 y + 1 = 2 z − 2 , then the vector equation of the line is A. r ⃗ = 1 3 i ^ − 1 3 j ^ + k ^ + λ ( i ^ + 2 j ^ + 3 k ^ ) \vec{r}=\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\hat{k}+\lambda(\hat{i}+2\hat{j}+3\hat{k}) r = 3 1 i ^ − 3 1 j ^ + k ^ + λ ( i ^ + 2 j ^ + 3 k ^ ) B. r ⃗ = ( i ^ + j ^ + k ^ ) + λ ( i ^ + 2 j ^ + 3 k ^ ) \vec{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda(\hat{i}+2\hat{j}+3\hat{k}) r = ( i ^ + j ^ + k ^ ) + λ ( i ^ + 2 j ^ + 3 k ^ ) C. r ⃗ = − 1 3 i ^ − 1 3 j ^ + k ^ + λ ( i ^ + 2 j ^ + 3 k ^ ) \vec{r}=-\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\hat{k}+\lambda(\hat{i}+2\hat{j}+3\hat{k}) r = − 3 1 i ^ − 3 1 j ^ + k ^ + λ ( i ^ + 2 j ^ + 3 k ^ ) D. r ⃗ = 1 3 i ^ − 1 3 j ^ − k ^ + λ ( i ^ − j ^ + 3 k ^ ) \vec{r}=\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}-\hat{k}+\lambda(\hat{i}-\hat{j}+3\hat{k}) r = 3 1 i ^ − 3 1 j ^ − k ^ + λ ( i ^ − j ^ + 3 k ^ ) Tap an option to check your answer.
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Question 4 11 May 2024 S1
147 The vector equation of the line 2 x + 4 = 3 y + 1 = 6 z − 3 2x+4=3y+1=6z-3 2 x + 4 = 3 y + 1 = 6 z − 3 is A. r ⃗ = ( − 2 i ^ + 1 3 j ^ + 1 2 k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) \vec{r}=(-2\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{2}\hat{k})+\lambda(3\hat{i}+2\hat{j}+\hat{k}) r = ( − 2 i ^ + 3 1 j ^ + 2 1 k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) B. r ⃗ = ( − 2 i ^ − 1 3 j ^ + 1 2 k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) \vec{r}=(-2\hat{i}-\frac{1}{3}\hat{j}+\frac{1}{2}\hat{k})+\lambda(3\hat{i}+2\hat{j}+\hat{k}) r = ( − 2 i ^ − 3 1 j ^ + 2 1 k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) C. r ⃗ = ( 2 i ^ + j ^ + k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) \vec{r}=(2\hat{i}+\hat{j}+\hat{k})+\lambda(3\hat{i}+2\hat{j}+\hat{k}) r = ( 2 i ^ + j ^ + k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) D. r ⃗ = ( − 2 i ^ − j ^ + k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) \vec{r}=(-2\hat{i}-\hat{j}+\hat{k})+\lambda(3\hat{i}+2\hat{j}+\hat{k}) r = ( − 2 i ^ − j ^ + k ^ ) + λ ( 3 i ^ + 2 j ^ + k ^ ) Tap an option to check your answer.
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Question 5 16 Apr 2026 S1
94 The vector equation of the line whose cartesian equations are x = 2 , 2 y − 3 z + 7 = 0 x = 2, 2y - 3z + 7 = 0 x = 2 , 2 y − 3 z + 7 = 0 A. r ⃗ = ( 2 i ^ + 2 j ^ − 3 k ^ ) + λ ( 2 j ^ − 3 k ^ ) \vec{r} = (2\hat{i} + 2\hat{j} - 3\hat{k}) + \lambda(2\hat{j} - 3\hat{k}) r = ( 2 i ^ + 2 j ^ − 3 k ^ ) + λ ( 2 j ^ − 3 k ^ ) B. r ⃗ = ( 2 i ^ + 2 j ^ − 3 k ^ ) + λ ( 2 i ^ − 3 j ^ ) \vec{r} = (2\hat{i} + 2\hat{j} - 3\hat{k}) + \lambda(2\hat{i} - 3\hat{j}) r = ( 2 i ^ + 2 j ^ − 3 k ^ ) + λ ( 2 i ^ − 3 j ^ ) C. r ⃗ = ( 2 i ^ + 7 3 k ^ ) + λ ( 3 j ^ + 2 k ^ ) \vec{r} = \left(2\hat{i} + \dfrac{7}{3}\hat{k}\right) + \lambda(3\hat{j} + 2\hat{k}) r = ( 2 i ^ + 3 7 k ^ ) + λ ( 3 j ^ + 2 k ^ ) D. r ⃗ = ( − 2 i ^ + 7 3 k ^ ) + λ ( 3 j ^ + 2 k ^ ) \vec{r} = \left(-2\hat{i} + \dfrac{7}{3}\hat{k}\right) + \lambda(3\hat{j} + 2\hat{k}) r = ( − 2 i ^ + 3 7 k ^ ) + λ ( 3 j ^ + 2 k ^ ) Tap an option to check your answer.
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Question 6 19 Apr 2026 S1
96 The symmetric form of the equation of the line x = a y + b x = ay + b x = a y + b , z = c y + d z = cy + d z = cy + d is A. x − a b = y = z − c d \dfrac{x - a}{b} = y = \dfrac{z - c}{d} b x − a = y = d z − c B. x − b a = y − d c = z \dfrac{x - b}{a} = \dfrac{y - d}{c} = z a x − b = c y − d = z C. x = y − a b = z − c d x = \dfrac{y - a}{b} = \dfrac{z - c}{d} x = b y − a = d z − c D. x − b a = y = z − d c \dfrac{x - b}{a} = y = \dfrac{z - d}{c} a x − b = y = c z − d Tap an option to check your answer.
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