Match List-I with List-II List-

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Vectors

Question:

Match List-I with List-II

List-I	List-II
(A) The vectors $λ\hat i+\hat j+2\hat k$ and $\hat i+\hat j+\hat k$ are perpendicular if $λ$ is equal to	(I) 1
(B) The vectors $3\hat i+6\hat j -\hat k$ and $2\hat i+4\hat j-λ\hat k$ are collinear if $λ$ is equal to	(II) -3
(C) The number of vectors of unit-length which are perpendicular to both the vectors $\vec a=\hat i+\hat j+2\hat k$ and $\vec b = 3\hat i-\hat j+5\hat k$ is	(III) 2/3
(D) If $\|\vec a\| = 1$ and $\vec a+\vec b=\vec 0$, then $\|\vec b\|$ is equal to	(IV) 2

Choose the correct answer from the options given below:

Options:

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

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

Explanation:

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

List-I	List-II
(A) The vectors $λ\hat i+\hat j+2\hat k$ and $\hat i+\hat j+\hat k$ are perpendicular if $λ$ is equal to	(II) -3
(B) The vectors $3\hat i+6\hat j -\hat k$ and $2\hat i+4\hat j-λ\hat k$ are collinear if $λ$ is equal to	(III) 2/3
(C) The number of vectors of unit-length which are perpendicular to both the vectors $\vec a=\hat i+\hat j+2\hat k$ and $\vec b = 3\hat i-\hat j+5\hat k$ is	(IV) 2
(D) If $\|\vec a\| = 1$ and $\vec a+\vec b=\vec 0$, then $\|\vec b\|$ is equal to	(I) 1

(A) For perpendicularity: $(\lambda,1,2)\cdot(1,1,1)=0\Rightarrow \lambda+1+2=0\Rightarrow \lambda=-3$

(B) Collinear: $\frac{3}{2}=\frac{6}{4}=\frac{-1}{-\lambda}\Rightarrow \frac{1}{\lambda}=\frac{3}{2}\Rightarrow \lambda=\frac{2}{3}$

(C) Vectors perpendicular to both $\vec a$ and $\vec b$ are along $\vec a\times\vec b\neq\vec 0$, so two unit vectors ($\pm$ direction) ⇒ $2$

(D) $|{\vec a}|=1,\ \vec a+\vec b=\vec 0\Rightarrow \vec b=-\vec a\Rightarrow |\vec b|=1$

The correct matching is A–II, B–III, C–IV, D–I.