Statement (A): If vec{a}, vec{b}, vec{c} are | vedclass

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

Question

(A) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = 1, |\vec{b}| = 1, |\vec{c}| = 1$.We are given $\vec{a} + \vec{b} + \v

Vedclass · Accepted Answer

$A$ and $R$ are both independently true and $R$ is the correct explanation for $A$.

Vedclass · Answer

(A) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = 1, |\vec{b}| = 1, |\vec{c}| = 1$.We are given $\vec{a} + \vec{b} + \vec{c} = 0$.Squaring both sides,we get $(\vec{a} + \vec{b} + \vec{c})^2 = 0^2 = 0$.Using the identity $(\vec{x} + \vec{y} + \vec{z})^2 = |\vec{x}|^2 + |\vec{y}|^2 + |\vec{z}|^2 + 2(\vec{x} \cdot \vec{y} + \vec{y} \cdot \vec{z} + \vec{z} \cdot \vec{x})$,we have:$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.Substituting the magnitudes,we get $1^2 + 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.$3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3$.Therefore,$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.Since the identity used in Reason $(R)$ is a fundamental property of the dot product and is used to derive the result in Statement $(A)$,$R$ is the correct explanation for $A$.

Statement $(A):$ If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$,then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.
Reason $(R): (\vec{x} + \vec{y})^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$.

Similar Questions

For any two vectors $\vec{a}$ and $\vec{b}$,which of the following statements is true?

If $|\overline{a}|=2, |\overline{b}|=3, |\overline{c}|=5$ and each of the angles between the vectors $\overline{a}$ and $\overline{b}$,$\overline{b}$ and $\overline{c}$,and $\overline{c}$ and $\overline{a}$ is $60^{\circ}$,then the value of $|\overline{a}+\overline{b}+\overline{c}|$ is

If $\hat{i}$ is the position vector of the centroid $G$ of triangle $ABC$ and $2\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}-4\hat{k}$ are respectively the position vectors of its vertices $A$ and $B$,then $AG^2+BG^2+CG^2=$

For any two vectors $\vec{a}$ and $\vec{b},$ we always have $|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$ (triangle inequality).

$A$ unit vector which is in the opposite direction of $(2, -2, 1)$ is $........$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Statement $(A):$ If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$,then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.Reason $(R): (\vec{x} + \vec{y})^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$.