If overline{b} and overline{c} are unit vecto | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Using the vector triple product formula $\bar{u} 	imes (\bar{v} 	imes \bar{w}) = (\bar{u} \cdot \bar{w})\bar{v} - (\bar{u} \cdot \bar{v})\bar{w}

Vedclass · Accepted Answer

$90^{\circ}, 120^{\circ}$

Vedclass · Answer

(C) Using the vector triple product formula $\bar{u} 	imes (\bar{v} 	imes \bar{w}) = (\bar{u} \cdot \bar{w})\bar{v} - (\bar{u} \cdot \bar{v})\bar{w}$,we expand the given expression:$\bar{a} 	imes (\bar{b} 	imes \bar{c}) = (\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}$$\bar{b} 	imes (\bar{c} 	imes \bar{a}) = (\bar{b} \cdot \bar{a})\bar{c} - (\bar{b} \cdot \bar{c})\bar{a}$Adding these,we get: $(\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c} + (\bar{b} \cdot \bar{a})\bar{c} - (\bar{b} \cdot \bar{c})\bar{a} = \frac{1}{2}\bar{a}$Since $(\bar{a} \cdot \bar{b})\bar{c}$ and $(\bar{b} \cdot \bar{a})\bar{c}$ cancel out,we have: $(\bar{a} \cdot \bar{c})\bar{b} - (\bar{b} \cdot \bar{c})\bar{a} = \frac{1}{2}\bar{a}$Rearranging: $(\bar{a} \cdot \bar{c})\bar{b} = (\frac{1}{2} + \bar{b} \cdot \bar{c})\bar{a}$Since $\bar{a}$ and $\bar{b}$ are not collinear,the coefficients must be zero: $\bar{a} \cdot \bar{c} = 0$ and $\frac{1}{2} + \bar{b} \cdot \bar{c} = 0$$\bar{a} \cdot \bar{c} = 0 \implies 	heta_{ac} = 90^{\circ}$$\bar{b} \cdot \bar{c} = -\frac{1}{2} \implies |\bar{b}||\bar{c}| \cos(	heta_{bc}) = -\frac{1}{2} \implies (1)(1) \cos(	heta_{bc}) = -\frac{1}{2} \implies 	heta_{bc} = 120^{\circ}$Thus,the angles are $90^{\circ}$ and $120^{\circ}$.

Similar Questions

Let $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}-2 \hat{k}$ and $\vec{c}=-\hat{i}+4 \hat{j}+3 \hat{k}$. If $\vec{d}$ is a vector perpendicular to both $\vec{b}$ and $\vec{c}$ and $\vec{a} \cdot \vec{d}=18$,then $|\vec{a} \times \vec{d}|^2$ is equal to $..........$.

Let $\vec{v}$ be a unit vector which follows the equation $\vec{v} \times \vec{b} = \vec{c}$. Also,$|\vec{b}| = 2$ and $|\vec{c}| = \sqrt{3}$. Then,which of the following is true?

If $a, b, c$ are three unit vectors such that $a \times (b \times c) = \frac{\sqrt{3}}{2} b + \frac{1}{2} c$,then the angles between $a, b$ and $a, c$ respectively are

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero vectors and $\hat{n}$ is a unit vector perpendicular to $\vec{c}$ such that $\vec{a} = \alpha \vec{b} - \hat{n}, (\alpha \neq 0)$ and $\vec{b} \cdot \vec{c} = 12$,then $|\vec{c} \times (\vec{a} \times \vec{b})|$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module