The cartesian equation of the line passing through the point (-4, 3, -2) and parallel to the line r = 5i - k + (i-j+2k) is: Option 1: (x-4)/1 = (y-3)/(-1) = (z-2)/2 Option 2: (x+4)/1 = (y-3)/(-1) = (z+2)/2 Option 3: (x+4)/5 = (y-3)/0 = (z+2)/(-1) Option 4: (x-4)/5 = (y-3)/0 = (z-2)/(-1)

Answer: (x+4)/1 = (y-3)/(-1) = (z+2)/2. Solution: A parallel line has the same direction ratios 1,-1,2. Through (-4,3,-2) the symmetric form is (x+4)/1=(y-3)/(-1)=(z+2)/2.

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The cartesian equation of the line passing through the point $(-4, 3, -2)$ and parallel to the line $\vec{r} = 5\hat{i} - \hat{k} + \lambda(\hat{i}-\hat{j}+2\hat{k})$ is:

$\dfrac{x-4}{1} = \dfrac{y-3}{-1} = \dfrac{z-2}{2}$

$\dfrac{x+4}{1} = \dfrac{y-3}{-1} = \dfrac{z+2}{2}$

$\dfrac{x+4}{5} = \dfrac{y-3}{0} = \dfrac{z+2}{-1}$

$\dfrac{x-4}{5} = \dfrac{y-3}{0} = \dfrac{z-2}{-1}$

✅ Correct Option: 2

A parallel line has the same direction ratios $\langle 1,-1,2\rangle$ .

Through $(-4,3,-2)$ the symmetric form is $\dfrac{x+4}{1}=\dfrac{y-3}{-1}=\dfrac{z+2}{2}$ .

2026: 30 May Shift 1

The shortest distance between lines $L_1:\frac{x+3}{-4}=\frac{6-y}{-3}=\frac{z}{2}$ and $L_2:\frac{x+2}{-4}=\frac{y}{1}=\frac{z-7}{1}$ is:

2025: 27 May Shift 1

Match List-I with List-II

List-I	List-II
(A) Line : $x = 2y + 1 = z - 1$	(I) Crosses $xz$ plane at (1, 0, 1)
(B) Line : $x + 1 = 2y + 1 = z$	(II) Crosses $xz$ plane at (0, 0, 1)
(C) Line : $x - 1 = 2y = z + 1$	(III) Crosses $xz$ plane at (1, 0, -1)
(D) Line : $x - 1 = 2y = z - 1$	(IV) Crosses $xz$ plane at (1, 0, 2)

Choose the correct answer from the options given below:

2022: 7 Aug Shift 2

The equation of plane that contains line $\frac{x - 1}{-1} = \frac{y + 1}{2} = \frac{z - 1}{3}$ and also pass through point $(0, 1, 0)$ is :

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The angle between the lines $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{j} - 2\hat{k} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$ is :

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The shortest distances of the point $(1, 2, 3)$ from $x$ , $y$ , $z$ axes respectively are

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If the points (-1, -1, 2), (2, m, 5) and (3, 11, 6) are collinear, then m equals

The cartesian equation of the line passing through the point (−4,3,−2)(-4, 3, -2)(−4,3,−2) and parallel to the line r⃗=5i^−k^+λ(i^−j^+2k^)\vec{r} = 5\hat{i} - \hat{k} + \lambda(\hat{i}-\hat{j}+2\hat{k})r=5i^−k^+λ(i^−j^​+2k^) is:

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CUET Mathematics Past Year Questions (Topic-Wise)

The cartesian equation of the line passing through the point $(-4, 3, -2)$ and parallel to the line $\vec{r} = 5\hat{i} - \hat{k} + \lambda(\hat{i}-\hat{j}+2\hat{k})$ is: