If a, b and c are three vectors such that a + b + c = 0, a, = 7, b, = 3 and, c, = 5, then angle between b and c is Option 1: (2)/3 Option 2: ()/6 Option 3: (5)/6 Option 4: ()/3

Answer: ()/3. Solution: Given a + b + c = 0, a, = 7, b, = 3, and, c, = 5. From the condition a + b + c = 0: a = -(b + c) Taking magnitude on both sides:, a, =, b + c, a, ^2 =, b + c, ^2 Using the dot product expansion, b + c, ^2 =, b, ^2 +, c, ^2 + 2b * c:, a, ^2 =, b, ^2 +, c, ^2 + 2b * c Substituting the given values: 7^2 = 3^2 + 5^2 + 2b * c 49 = 9 + 25 + 2b * c 49 = 34 + 2b * c 15 = 2b * c b * c = 15/2 Using the dot product formula b * c =, b, x, c, x, where is the angle between b and c: 15/2 = 3 x 5 x 15/2 = 15 = 1/2 Therefore: = 60° = ()/3 The angle between b and c is ()/3.

Algebra

Vector Algebra

Medium

core

If $\vec{a}, \vec{b}$ and $\vec{c}$ are three vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , $|\vec{a}| = 7$ , $|\vec{b}| = 3$ and $|\vec{c}| = 5$ , then angle between $\vec{b}$ and $\vec{c}$ is

$\frac{2\pi}{3}$

$\frac{\pi}{6}$

$\frac{5\pi}{6}$

$\frac{\pi}{3}$

✅ Correct Option: 4

Given $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , $|\vec{a}| = 7$ , $|\vec{b}| = 3$ , and $|\vec{c}| = 5$ .

From the condition $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ :

$\vec{a} = -(\vec{b} + \vec{c})$

Taking magnitude on both sides:

$|\vec{a}| = |\vec{b} + \vec{c}|$

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

Using the dot product expansion $|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}$ :

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

Substituting the given values:

$7^2 = 3^2 + 5^2 + 2\vec{b} \cdot \vec{c}$

$49 = 9 + 25 + 2\vec{b} \cdot \vec{c}$

$49 = 34 + 2\vec{b} \cdot \vec{c}$

$15 = 2\vec{b} \cdot \vec{c}$

$\vec{b} \cdot \vec{c} = \frac{15}{2}$

Using the dot product formula $\vec{b} \cdot \vec{c} = |\vec{b}| \times |\vec{c}| \times \cos\theta$ , where $\theta$ is the angle between $\vec{b}$ and $\vec{c}$ :

$\frac{15}{2} = 3 \times 5 \times \cos\theta$

$\frac{15}{2} = 15\cos\theta$

$\cos\theta = \frac{1}{2}$

Therefore:

$\theta = 60° = \frac{\pi}{3}$

The angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$ .

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For the vector $\vec a = \dfrac{1}{3}(2\hat i - 2\hat j + \hat k)$ , which of the following statements is/are TRUE ?

A. vector $\vec a$ is a unit vector

B. vector $\vec a$ is making an angle $\dfrac{\pi}{3}$ with vector $2\hat i - 4\hat j + 3\hat k$

C. vector $\vec a$ is parallel to vector $-\hat i + \hat j - \dfrac{1}{2}\hat k$

D. vector $\vec a$ is perpendicular to vector $3\hat i + 2\hat j - 2\hat k$

Choose the correct answer from the options given below:

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In $\triangle ABC$ , Which of the following are True?

A. $\overline{AB}+\overline{BC}+\overline{CA}=\vec{0}$

B. $\overline{AB}+\overline{BC}-\overline{AC}=\vec{0}$

C. $\overline{AB}+\overline{BC}-\overline{CA}=\vec{0}$

D. $\overline{AB}-\overline{CB}+\overline{CA}=\vec{0}$

Choose the correct answer from the options given below:

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If $\theta$ is the angle between two unit vectors $\hat{a}$ and $\hat{b}$ then $|\hat{a}-\hat{b}| =$

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If $|\vec{a}| = 3$ and $|\vec{b}| = 4$ , then a value of $\lambda$ for which $\vec{a} + \lambda \vec{b}$ and $\vec{a} - \lambda \vec{b}$ are perpendicular is :

Solution

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