Let, a, = 5, b, = 2 and a*b = 6, then the value of, a x b, is equal to Option 1: 8 Option 2: 6 Option 3: 10 Option 4: 60

Answer: 8. Solution: Given, a, = 5, b, = 2, and ab = 6 Using the dot product formula ab =, a, b, : 6 = 5 x 2 x 6 = 10 = 6/10 = 0.6 Using the identity ^2 + ^2 = 1: ^2 = 1 - ^2 ^2 = 1 - (0.6)^2 ^2 = 1 - 0.36 ^2 = 0.64 = 0.8 Note: The angle between two vectors is always between 0° and 180°, so is positive. Using the cross product magnitude formula, a x b, =, a, b, :, a x b, = 5 x 2 x 0.8, a x b, = 10 x 0.8, a x b, = 8 Therefore, a x b, = 8

Algebra

Vector Algebra

Medium

core

Let $|\vec{a}| = 5, |\vec{b}| = 2$ and $\vec{a}\cdot\vec{b} = 6$ , then the value of $|\vec{a} \times \vec{b}|$ is equal to

✅ Correct Option: 1

Given $|\vec{a}| = 5$ , $|\vec{b}| = 2$ , and $\vec{a}\cdot\vec{b} = 6$

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

$6 = 5 \times 2 \times \cos\theta$

$6 = 10\cos\theta$

$\cos\theta = \frac{6}{10} = 0.6$

Using the identity $\sin^2\theta + \cos^2\theta = 1$ :

$\sin^2\theta = 1 - \cos^2\theta$

$\sin^2\theta = 1 - (0.6)^2$

$\sin^2\theta = 1 - 0.36$

$\sin^2\theta = 0.64$

$\sin\theta = 0.8$

Note: The angle between two vectors is always between $0°$ and $180°$ , so $\sin\theta$ is positive.

Using the cross product magnitude formula $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$ :

$|\vec{a} \times \vec{b}| = 5 \times 2 \times 0.8$

$|\vec{a} \times \vec{b}| = 10 \times 0.8$

$|\vec{a} \times \vec{b}| = 8$

Therefore, $|\vec{a} \times \vec{b}| = 8$

2026: 30 May Shift 2

If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined at an angle $\theta$ , then the value of $|\vec{a}-\vec{b}|$ is

2023: 15 June Shift 2

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = -2\hat{i} + \hat{j} - 2\hat{k}$ . Then

(A) $\vec{a}$ is a unit vector

(B) $\vec{a} \times \vec{b} = -\hat{i} + 2\hat{j} + 2\hat{k}$

(D) $\vec{a}$ and $\vec{b}$ are neither parallel nor perpendicular vectors

Choose the correct answer from the options given below :

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If $\vec{a}$ is a unit vector and $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 8$ then $|\vec{x}|$ is

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The vector $\{(a^2+a-2)\hat{i}+(a^2-4)\hat{j}-(a^2-5a+6)\hat{k}\}$ makes acute angles with coordinate axes, then values of $a$ lies in :

2026: 15 May Shift 1

If $|\vec a|=2, |\vec b|=7, \vec a \times \vec b = 3\hat i + 2\hat j + 6\hat k$ , then the angle between $\vec a$ and $\vec b$ is equal to

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If the angle between the vectors $\vec a=2\lambda^2\hat i+4\lambda\hat j+\hat k$ and $\vec b=7\hat i-2\hat j+\lambda\hat k$ is obtuse, then

Let ∣a⃗∣=5,∣b⃗∣=2|\vec{a}| = 5, |\vec{b}| = 2∣a∣=5,∣b∣=2 and a⃗⋅b⃗=6\vec{a}\cdot\vec{b} = 6a⋅b=6, then the value of ∣a⃗×b⃗∣|\vec{a} \times \vec{b}|∣a×b∣ is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Let $|\vec{a}| = 5, |\vec{b}| = 2$ and $\vec{a}\cdot\vec{b} = 6$ , then the value of $|\vec{a} \times \vec{b}|$ is equal to