Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

Match List-I with List-II

List-I

List-II

(A) Line: $x = 2y+1=z-1$

(1) 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:

Options:

(A)-(IV), (B)-(II), (C)-(III), (D)-(I)

(A)-(II), (B)-(IV), (C)-(III), (D)-(I)

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Correct Answer:

(A)-(IV), (B)-(II), (C)-(III), (D)-(I)

Explanation:

The correct answer is Option (1) → (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

To find where a line crosses the xz-plane, set y = 0 in the line's equation and solve for x and z.

(A) Line: x = 2y + 1 = z − 1

Set y = 0 → x = 1, z = x + 1 = 2 → point = (1, 0, 2)

⇒ Matches with (IV)

(B) Line: x + 1 = 2y + 1 = z

Set y = 0 → 2y + 1 = 1 → x + 1 = 1 ⇒ x = 0, z = 1 → point = (0, 0, 1)

⇒ Matches with (II)

(C) Line: x − 1 = 2y = z + 1

Set y = 0 → 2y = 0 ⇒ x − 1 = 0 ⇒ x = 1, z + 1 = 0 ⇒ z = −1 → point = (1, 0, -1)

⇒ Matches with (III)

(D) Line: x − 1 = 2y = z − 1

Set y = 0 → 2y = 0 ⇒ x − 1 = 0 ⇒ x = 1, z − 1 = 0 ⇒ z = 1 → point = (1, 0, 1)

⇒ Matches with (I)