If a, b and √3a - b are three unit vectors, then the angle between a and b is: Option 1: ()/6 Option 2: ()/4 Option 3: ()/3 Option 4: ()/2

Answer: ()/6. Solution: Given a, b, and √3a - b are three unit vectors. Since a and b are unit vectors:, a, = 1 and, b, = 1 Since √3a - b is also a unit vector:, √3a - b, = 1, √3a - b, ^2 = 1 (√3a - b) * (√3a - b) = 1 (√3)^2(a * a) - 2√3(a * b) + (b * b) = 1 3(a * a) - 2√3(a * b) + (b * b) = 1 Since a and b are unit vectors: a * a =, a, ^2 = 1 b * b =, b, ^2 = 1 Therefore: 3(1) - 2√3(a * b) + 1 = 1 4 - 2√3(a * b) = 1 3 = 2√3(a * b) a * b = 3/2√3 a * b = 3/2√3 x √3/√3 a * b = 3√3/6 a * b = √3/2 Using the dot product formula a * b =, a, b, : √3/2 = (1)(1) = √3/2 = ()/6 Therefore, the angle between a and b is ()/6.

Algebra

Vector Algebra

Medium

core

If $\vec{a}$ , $\vec{b}$ and $\sqrt{3}\vec{a} - \vec{b}$ are three unit vectors, then the angle between $\vec{a}$ and $\vec{b}$ is:

$\frac{\pi}{6}$

$\frac{\pi}{4}$

$\frac{\pi}{3}$

$\frac{\pi}{2}$

✅ Correct Option: 1

Given $\vec{a}$ , $\vec{b}$ , and $\sqrt{3}\vec{a} - \vec{b}$ are three unit vectors.

Since $\vec{a}$ and $\vec{b}$ are unit vectors:

$|\vec{a}| = 1$ and $|\vec{b}| = 1$

Since $\sqrt{3}\vec{a} - \vec{b}$ is also a unit vector:

$|\sqrt{3}\vec{a} - \vec{b}| = 1$

$|\sqrt{3}\vec{a} - \vec{b}|^2 = 1$

$(\sqrt{3}\vec{a} - \vec{b}) \cdot (\sqrt{3}\vec{a} - \vec{b}) = 1$

$(\sqrt{3})^2(\vec{a} \cdot \vec{a}) - 2\sqrt{3}(\vec{a} \cdot \vec{b}) + (\vec{b} \cdot \vec{b}) = 1$

$3(\vec{a} \cdot \vec{a}) - 2\sqrt{3}(\vec{a} \cdot \vec{b}) + (\vec{b} \cdot \vec{b}) = 1$

Since $\vec{a}$ and $\vec{b}$ are unit vectors:

$\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1$

$\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1$

Therefore:

$3(1) - 2\sqrt{3}(\vec{a} \cdot \vec{b}) + 1 = 1$

$4 - 2\sqrt{3}(\vec{a} \cdot \vec{b}) = 1$

$3 = 2\sqrt{3}(\vec{a} \cdot \vec{b})$

$\vec{a} \cdot \vec{b} = \frac{3}{2\sqrt{3}}$

$\vec{a} \cdot \vec{b} = \frac{3}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\vec{a} \cdot \vec{b} = \frac{3\sqrt{3}}{6}$

$\vec{a} \cdot \vec{b} = \frac{\sqrt{3}}{2}$

Using the dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$ :

$\frac{\sqrt{3}}{2} = (1)(1)\cos\theta$

$\cos\theta = \frac{\sqrt{3}}{2}$

$\theta = \frac{\pi}{6}$

Therefore, the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$ .

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D. The position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2) is $3\hat{i}+2\hat{j}-\hat{k}$ .

Choose the correct answer from the options given below:

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Let $\vec{a},\vec{b},\vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=2, |\vec{c}|=1$ . If the projection of $\vec{b}$ along $\vec{c}$ is equal to the projection of $\vec{a}$ along $\vec{c}$ and $\vec{b} \perp \vec{a}$ , then the value of $|2\vec{a}-2\vec{b}+3\vec{c}|$ is equal to

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If $\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} - 3\hat{j} - 5\hat{k}$ then $(\vec{a}+\vec{b}).(\vec{a}-\vec{b})$ is equal to:

If a⃗\vec{a}a, b⃗\vec{b}b and 3a⃗−b⃗\sqrt{3}\vec{a} - \vec{b}3​a−b are three unit vectors, then the angle between a⃗\vec{a}a and b⃗\vec{b}b is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $\vec{a}$ , $\vec{b}$ and $\sqrt{3}\vec{a} - \vec{b}$ are three unit vectors, then the angle between $\vec{a}$ and $\vec{b}$ is: