(vec{a}+vec{b}) cdot (vec{a}+vec{b}) = |vec{a | vedclass

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

Question

(B) Given the equation: $(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$. Expanding the left side using the distributive proper

Vedclass · Accepted Answer

$\vec{a}$ and $\vec{b}$ are perpendicular to each other.

Vedclass · Answer

(B) Given the equation: $(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$. Expanding the left side using the distributive property of the dot product: $\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = |\vec{a}|^2 + |\vec{b}|^2$. Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2$,and the dot product is commutative $(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a})$,we get: $|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2$. Subtracting $|\vec{a}|^2 + |\vec{b}|^2$ from both sides: $2(\vec{a} \cdot \vec{b}) = 0$. This implies $\vec{a} \cdot \vec{b} = 0$. Since $\vec{a} 
eq \vec{0}$ and $\vec{b} 
eq \vec{0}$,the dot product of two non-zero vectors is zero if and only if they are perpendicular to each other. Thus,the correct option is $B$.

$(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$ if and only if . . . . . . (where $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$).

Similar Questions

If the points $P$ and $Q$ are respectively the circumcentre and the orthocentre of a $\triangle ABC$,then $\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}$ is equal to

Let $\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}$,$\vec{b}=4 \hat{i}+\hat{j}$,$\vec{c}=\hat{i}-3 \hat{j}-7 \hat{k}$. If $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$,$\vec{r} \cdot \vec{a}=9$,$\vec{r} \cdot \vec{b}=7$,$\vec{r} \cdot \vec{c}=6$,then $(x, y, z) = $

If $p \times q = p \times r$ and $p \cdot q = p \cdot r$,then $\ldots . . .$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module