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Home Questions 8PDD2JFZ87
Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

Match List-I with List-II

List-I Equation of line

List-II Information

(A) $\vec r= (3\hat i-2\hat j+\hat k) + λ(\hat j-2\hat k)$

(I) Direction ratios are 2, 4, -1

(B) $\frac{2-x}{1}=\frac{2y+1}{4},z=2$

(II) Perpendicular to $2\hat i - \hat j +\hat k$

(C) $\frac{x}{1}=\frac{y-3}{2}=\frac{3-4z}{2}$

(III) Passing through the point (3, -2, 1)

(D) $\vec r= (3\hat i+2\hat j + \hat k) + λ(2\hat i+\hat j −3\hat k)$

(IV) Direction ratios are -1, 2, 0

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

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

List-I Equation of line

List-II Information

(A) $\vec r= (3\hat i-2\hat j+\hat k) + λ(\hat j-2\hat k)$

(III) Passing through the point (3, -2, 1)

(B) $\frac{2-x}{1}=\frac{2y+1}{4},z=2$

(IV) Direction ratios are -1, 2, 0

(C) $\frac{x}{1}=\frac{y-3}{2}=\frac{3-4z}{2}$

(I) Direction ratios are 2, 4, -1

(D) $\vec r= (3\hat i+2\hat j + \hat k) + λ(2\hat i+\hat j −3\hat k)$

(II) Perpendicular to $2\hat i - \hat j +\hat k$

(A) $\vec r=(3\hat i-2\hat j+\hat k)+\lambda(\hat j-2\hat k)$ has DR $(0,1,-2)$ and passes through $(3,-2,1)$ → (III).

(B) $\frac{2-x}{1}=\frac{2y+1}{4},\ z=2$ ⇒ let $t$: $x=2-t,\ y=\frac{4t-1}{2},\ z=2$. DR $(-1,2,0)$ → (IV).

(C) $\frac{x}{1}=\frac{y-3}{2}=\frac{3-4z}{2}$ ⇒ $x=t,\ y=3+2t,\ z=\frac{3-2t}{4}$. DR $(1,2,-\frac12)\propto(2,4,-1)$ → (I).

(D) $\vec r=(3\hat i+2\hat j+\hat k)+\lambda(2\hat i+\hat j-3\hat k)$ has DR $(2,1,-3)$ and $(2,1,-3)\cdot(2,-1,1)=4-1-3=0$ ⇒ ⟂ to $2\hat i-\hat j+\hat k$ → (II).

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