Let vec{a}, vec{b} and vec{c} be three non-ze | vedclass

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

Question

(B) Given the vector triple product identity: $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the vector triple product identity: $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$.Equating this to the given expression: $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = \frac{1}{3}|\vec{b}| |\vec{c}| \vec{a}$.Since $\vec{a}, \vec{b}, \vec{c}$ are non-zero and no two are collinear,the vectors $\vec{a}$ and $\vec{b}$ are linearly independent. For the equation to hold,the coefficients of $\vec{a}$ and $\vec{b}$ must be zero.Thus,$\vec{a} \cdot \vec{c} = 0$ and $-(\vec{b} \cdot \vec{c}) = \frac{1}{3}|\vec{b}| |\vec{c}|$.Using the definition of the dot product $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos 	heta$,we get:$-|\vec{b}| |\vec{c}| \cos 	heta = \frac{1}{3}|\vec{b}| |\vec{c}|$.Dividing by $|\vec{b}| |\vec{c}|$ (as they are non-zero),we find $\cos 	heta = -\frac{1}{3}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin^2 	heta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$.Therefore,$\sin 	heta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$ (since $	heta \in [0, \pi]$,$\sin 	heta \ge 0$).

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinear and $(\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3}|\vec{b}| |\vec{c}| \vec{a}$. If $\theta$ is the angle between vectors $\vec{b}$ and $\vec{c}$,then a value of $\sin \theta$ is:

Similar Questions

If $\overline{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\overline{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$,then the value of $(\overline{a}-2 \overline{b}) \cdot \{(\overline{a} \times \overline{b}) \times (2 \overline{a}+\overline{b})\}$ is

Let $\overline{a}=\hat{j}-\hat{k}$ and $\overline{c}=\hat{i}-\hat{j}-\hat{k}$. Then the vector $\overline{b}$ satisfying $\overline{a} \times \overline{b}+\overline{c}=\overline{0}$ and $\overline{a} \cdot \overline{b}=3$,is

Let $\vec{a}, \vec{b},$ and $\vec{c}$ be three unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2}(\vec{b} + \vec{c})$. If $\vec{b}$ is not parallel to $\vec{c}$,then the angle between $\vec{a}$ and $\vec{b}$ is:

If $a = i + j - 2k$, then $\sum \{(a \times i) \times j\}^2$ is equal to

If $a \cdot b = \beta$ and $a \times b = c$, then $a =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module