If a(vec{alpha} times vec{beta}) + b(vec{beta | vedclass

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

Question

(C) Given the equation $a(\vec{\alpha} 	imes \vec{\beta}) + b(\vec{\beta} 	imes \vec{\gamma}) + c(\vec{\gamma} 	imes \vec{\alpha}) = \overrightarro

Vedclass · Accepted Answer

coplanar

Vedclass · Answer

(C) Given the equation $a(\vec{\alpha} 	imes \vec{\beta}) + b(\vec{\beta} 	imes \vec{\gamma}) + c(\vec{\gamma} 	imes \vec{\alpha}) = \overrightarrow{0}$.Let $\vec{u} = \vec{\alpha} 	imes \vec{\beta}$, $\vec{v} = \vec{\beta} 	imes \vec{\gamma}$, and $\vec{w} = \vec{\gamma} 	imes \vec{\alpha}$.The given equation is $a\vec{u} + b\vec{v} + c\vec{w} = \overrightarrow{0}$.Since $a, b, c$ are non-zero scalars, the vectors $\vec{u}, \vec{v}, \vec{w}$ are linearly dependent, which implies they are coplanar.For any three vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$, the cross products $\vec{\alpha} 	imes \vec{\beta}$, $\vec{\beta} 	imes \vec{\gamma}$, and $\vec{\gamma} 	imes \vec{\alpha}$ are coplanar if and only if $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are coplanar.Therefore, the vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are coplanar.

If $a(\vec{\alpha} \times \vec{\beta}) + b(\vec{\beta} \times \vec{\gamma}) + c(\vec{\gamma} \times \vec{\alpha}) = \overrightarrow{0}$, where $a, b, c$ are non-zero scalars, then the vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are

Similar Questions

The vector $c$ directed along the internal bisector of the angle between the vectors $a = 7i - 4j - 4k$ and $b = -2i - j + 2k$ with $|c| = 5\sqrt{6}$ is

If $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$ and $\vec{c} = \hat{i} + 3\hat{j} - \hat{k}$,find $\lambda$ such that $\vec{a}$ is perpendicular to $\lambda\vec{b} + \vec{c}$.

For a triangle $ABC$ with vertices $A(1, 0, 0)$,$B(0, 1, 0)$,and $C(0, 0, 1)$,the angle $A = \dots$

If $a, b, c, d$ are the position vectors of the points $A, B, C$ and $D$ respectively,referred to the same origin $O$,such that no three of these points are collinear and $a + c = b + d$,then the quadrilateral $ABCD$ is a

Find $|\vec{a}|$ and $|\vec{b}|,$ if $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module