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CUET Mathematics 2026 PYQs31 May Shift 2. Free, no login required.

Geometry

past year questions, CUET Mathematics 2026 31 May Shift 2

3D Geometry

past year questions, Geometry, CUET Mathematics 2026 31 May Shift 2

Medium

core

If the lines x=ay+bx = ay+b, z=cy+dz = cy+d and x=py+qx = py+q, z=ry+sz = ry+s are perpendicular, then ap+crap + cr is equal to

Solution

✅ Correct Option: 3

Writing each line in symmetric form with yy as parameter, the direction ratios are (a,1,c)(a,1,c) and (p,1,r)(p,1,r). Perpendicularity gives ap+1+cr=0ap + 1 + cr = 0, so ap+cr=−1ap + cr = -1.

Related questions:

2026: 21 May Shift 2

Let L1L_1 and L2L_2 be any two lines given by

L1:3i^+2j^−4k^+λ(i^+2j^+2k^)L_1: 3\hat{i}+2\hat{j}-4\hat{k} + \lambda(\hat{i}+2\hat{j}+2\hat{k})

L2:5i^−2j^+μ(3i^+2j^+6k^)L_2: 5\hat{i}-2\hat{j} + \mu(3\hat{i}+2\hat{j}+6\hat{k}) Where λ\lambda and μ\mu are parameters

then which of the following statements are TRUE?

A. Equation of a line passing through the point (1,2,3)(1,2,3) which is parallel to line L1L_1 is x−11=y−22=z−32\dfrac{x-1}{1} = \dfrac{y-2}{2} = \dfrac{z-3}{2}

B. Direction cosines of line L2L_2 are 37,27,67\dfrac{3}{7}, \dfrac{2}{7}, \dfrac{6}{7}

C. Line L1L_1 is parallel to the line L2L_2

D. The angle between the two lines L1L_1 and L2L_2 is cos⁡−1(1921)\cos^{-1}\left(\dfrac{19}{21}\right)

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2026: 29 May Shift 2

If λ\lambda and μ\mu are parameters, then

Match the LIST-I with LIST-II

LIST-ILIST-II
A.r⃗=2i^+j^−3k^+λ(i^−j^+2k^)\vec{r}=2\hat{i}+\hat{j}-3\hat{k}+\lambda(\hat{i}-\hat{j}+2\hat{k}) parallel to the lineI. r⃗=(j^+k^)+μ(−2i^+j^−2k^)\vec{r}=(\hat{j}+\hat{k})+\mu(-2\hat{i}+\hat{j}-2\hat{k})
B.r⃗=i^+j^+λ(i^+j^−2k^)\vec{r}=\hat{i}+\hat{j}+\lambda(\hat{i}+\hat{j}-2\hat{k}) perpendicular to the lineII. r⃗=(2i^+k^)+μ(−2i^+j^)\vec{r}=(2\hat{i}+\hat{k})+\mu(-2\hat{i}+\hat{j})
C.x−2=y−11=z−1−2\frac{x}{-2}=\frac{y-1}{1}=\frac{z-1}{-2} in vector formIII. r⃗=(i^−j^+2k^)+μ(2i^−2j^+4k^)\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(2\hat{i}-2\hat{j}+4\hat{k})
D.2−x2=y,z=1\frac{2-x}{2}=y, z=1 in vector formIV. x−12=y+14=z3\frac{x-1}{2}=\frac{y+1}{4}=\frac{z}{3}

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2026: 21 May Shift 1

Match List-I with List-II

List-I (Straight Lines)List-II (Direction Ratios)
(A) r⃗=(i^+j^)+λ(2i^−j^+3k^)\vec{r} = (\hat{i}+\hat{j}) + \lambda(2\hat{i}-\hat{j}+3\hat{k})(I) 2,1,-3
(B) x2=1−y−1=z+13\frac{x}{2} = \frac{1-y}{-1} = \frac{z+1}{3}(II) -2,1,3
(C) 1−2x4=y+11=2z+36\frac{1-2x}{4} = \frac{y+1}{1} = \frac{2z+3}{6}(III) 2,-1,3
(D) r⃗=(2i^−k^)+λ(2i^+j^−3k^)\vec{r} = (2\hat{i}-\hat{k}) + \lambda(2\hat{i}+\hat{j}-3\hat{k})(IV) 2,1,3

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2026: 30 May Shift 2

Direction ratios of the line perpendicular to the lines x−32=y+7−3=z−21\frac{x-3}{2}=\frac{y+7}{-3}=\frac{z-2}{1} and x+21=y+32=z−5−2\frac{x+2}{1}=\frac{y+3}{2}=\frac{z-5}{-2} are

2025: 22 May Shift 2

Match List-I with List-II

List-IList-II
Equation of lineInformation
(A) r⃗=(3i^−2j^+k^)+λ(j^−2k^)\vec{r} = (3\hat{i} - 2\hat{j} + \hat{k}) + \lambda(\hat{j} - 2\hat{k})(I) Direction ratios are 2, 4, -1
(B) 2−x1=2y+14\frac{2-x}{1} = \frac{2y+1}{4}, z=2z = 2(II) Perpendicular to 2i^−j^+k^2\hat{i} - \hat{j} + \hat{k}
(C) x1=y−32=3−4z2\frac{x}{1} = \frac{y-3}{2} = \frac{3-4z}{2}(III) Passing through the point (3,−2,1)(3, -2, 1)
(D) r⃗=(3i^+2j^+k^)+λ(2i^+j^−3k^)\vec{r} = (3\hat{i} + 2\hat{j} + \hat{k}) + \lambda(2\hat{i} + \hat{j} - 3\hat{k})(IV) Direction ratios are -1, 2, 0

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2022: 6 Aug Shift 2

If two lines x−32=y−45=z4\frac{x-3}{2} = \frac{y-4}{5} = \frac{z}{4} and x−43=y−56=1−zk\frac{x-4}{3} = \frac{y-5}{6} = \frac{1-z}{k}, are coplanar, then kk is equal to

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