If, a, = 1, b, = 2, 2a+b, = 2√3 then, a-b, is: Option 1: √3 Option 2: √5 Option 3: 2/√3 Option 4: 2/√5

Answer: √3. Solution: Given, a, = 1, b, = 2, and, 2a+b, = 2√3. Starting with, 2a+b, = 2√3:, 2a+b, ^2 = (2√3)^2, 2a+b, ^2 = 12 Expanding the left side using the dot product: (2a+b) * (2a+b) = 12 4a * a + 4a * b + b * b = 12 Substituting a * a =, a, ^2 = 1 and b * b =, b, ^2 = 4: 4(1) + 4a * b + 4 = 12 8 + 4a * b = 12 4a * b = 4 a * b = 1 To find, a-b, :, a-b, ^2 = (a-b) * (a-b), a-b, ^2 = a * a - 2a * b + b * b, a-b, ^2 =, a, ^2 - 2a * b +, b, ^2, a-b, ^2 = 1 - 2(1) + 4, a-b, ^2 = 3, a-b, = √3

Algebra

Vector Algebra

Medium

core

If $|\vec{a}| = 1$ , $|\vec{b}| = 2$ , $|2\vec{a}+\vec{b}| = 2\sqrt{3}$ then $|\vec{a}-\vec{b}|$ is:

$\sqrt{3}$

$\sqrt{5}$

$\frac{2}{\sqrt{3}}$

$\frac{2}{\sqrt{5}}$

✅ Correct Option: 1

Given $|\vec{a}| = 1$ , $|\vec{b}| = 2$ , and $|2\vec{a}+\vec{b}| = 2\sqrt{3}$ .

Starting with $|2\vec{a}+\vec{b}| = 2\sqrt{3}$ :

$|2\vec{a}+\vec{b}|^2 = (2\sqrt{3})^2$

$|2\vec{a}+\vec{b}|^2 = 12$

Expanding the left side using the dot product:

$(2\vec{a}+\vec{b}) \cdot (2\vec{a}+\vec{b}) = 12$

$4\vec{a} \cdot \vec{a} + 4\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 12$

Substituting $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 4$ :

$4(1) + 4\vec{a} \cdot \vec{b} + 4 = 12$

$8 + 4\vec{a} \cdot \vec{b} = 12$

$4\vec{a} \cdot \vec{b} = 4$

$\vec{a} \cdot \vec{b} = 1$

To find $|\vec{a}-\vec{b}|$ :

$|\vec{a}-\vec{b}|^2 = (\vec{a}-\vec{b}) \cdot (\vec{a}-\vec{b})$

$|\vec{a}-\vec{b}|^2 = \vec{a} \cdot \vec{a} - 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}$

$|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{b}|^2$

$|\vec{a}-\vec{b}|^2 = 1 - 2(1) + 4$

$|\vec{a}-\vec{b}|^2 = 3$

$|\vec{a}-\vec{b}| = \sqrt{3}$

2025: 14 May Shift 1

If $\hat{i},\hat{j}$ and $\hat{k}$ are unit vectors along co-ordinates axes OX, OY and OZ respectively, then which of the following is/are true?

(A) $\hat{i} \times \hat{i} = \vec{0}$

(B) $\hat{i} \times \hat{k} = \hat{j}$

(D) $\hat{i} \cdot \hat{j} = 0$

Choose the correct answer from the options given below:

2022: 6 Aug Shift 2

In a triangle, $\triangle ABC$ , the sides AB and AC are represented by vectors $\hat{i} + \hat{j} + \hat{k}$ and $2\hat{i} - \hat{k}$ respectively. The length of median drawn from vertex A to BC is:

2026: 22 May Shift 1

If $\vec a$ and $\vec b$ are two unit vectors and $|\vec a-\vec b|=\sqrt3$ , then the value of $|\vec a+\vec b|$ is:

2022: 17 Aug Shift 2

Arrange the vectors in descending order of their magnitudes.

(A) $\hat{i} + \hat{j} + \hat{k}$

(B) $2\hat{i} - 3\hat{j}$

(D) $2\hat{i} - \hat{k}$

Choose the correct answer from the options given below :

2024: 16 May Shift 1

If $\vec{a}, \vec{b}$ and $\vec{c}$ are three vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$ , where $\vec{a}$ and $\vec{b}$ are unit vectors and $|\vec{c}|=2$ , then the angle between the vectors $\vec{b}$ and $\vec{c}$ is :

2026: 30 May Shift 1

Which of the following statements are correct ?

A. If $\vec{a}$ and $\vec{b}$ are two adjacent sides of a triangle, then the area of the triangle is $\frac{1}{2}|\vec{a}\times\vec{b}|$

B. If $\vec{a}$ and $\vec{b}$ are two adjacent sides of a triangle, then the area of the triangle is $\frac{1}{2}|\vec{a}.\vec{b}|$

C. If $\vec{a}$ and $\vec{b}$ are two adjacent sides of a parallelogram, then its area is $|\vec{a}\times\vec{b}|$

D. If $\vec{a}$ and $\vec{b}$ are two adjacent sides of a parallelogram, then its area is $|\vec{a}.\vec{b}|$

Choose the correct answer from the options given below:

If ∣a⃗∣=1|\vec{a}| = 1∣a∣=1, ∣b⃗∣=2|\vec{b}| = 2∣b∣=2, ∣2a⃗+b⃗∣=23|2\vec{a}+\vec{b}| = 2\sqrt{3}∣2a+b∣=23​ then ∣a⃗−b⃗∣|\vec{a}-\vec{b}|∣a−b∣ is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $|\vec{a}| = 1$ , $|\vec{b}| = 2$ , $|2\vec{a}+\vec{b}| = 2\sqrt{3}$ then $|\vec{a}-\vec{b}|$ is: