Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Vectors

Question:

Let $\vec a =\hat i + 2\hat j + 3\hat k$ and $\vec b=-2\hat i+ 3\hat j - 4\hat k$, then which of the following statements are correct?

(A) $|\vec a| = \sqrt{14}$
(B) $|\vec b| = 29$
(C) $\vec a-\vec b= 8$
(D) Angle between $\vec a$ and $\vec b$ is $\cos^{-1}(\frac{-8}{406})$

Choose the correct answer from the options given below:

Options:

(A) and (D) only

(A) and (C) only

(B), (C) and (D) only

(A), (C) and (D) only

Correct Answer:

(A) and (D) only

Explanation:

The correct answer is Option (1) → (A) and (D) only

Given:

$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$

$\vec{b} = -2\hat{i} + 3\hat{j} - 4\hat{k}$

(A) $|\vec{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$ ✅

(B) $|\vec{b}| = \sqrt{(-2)^2 + 3^2 + (-4)^2} = \sqrt{29} \ne 29$ ❌

(C) $\vec{a} \cdot \vec{b} = (1)(-2) + (2)(3) + (3)(-4) = -2 + 6 - 12 = -8$

Hence, $\vec{a} \cdot \vec{b} = -8 \ne 8$ ❌

(D) $\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{-8}{\sqrt{14}\sqrt{29}} = \frac{-8}{\sqrt{406}}$ ✅

Correct statements: (A), (D)