(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 $a \cdot i = a \cdot (i + j) = a \cdot (i + j + k)$, then $a = $

If the scalar projection of the vector $xi - j + k$ on the vector $2i - j + 5k$ is $\frac{1}{\sqrt{30}}$, then the value of $x$ is equal to

Let the unit vectors $a$ and $b$ be perpendicular and the unit vector $c$ be inclined at an angle $\theta$ to both $a$ and $b$. If $c = \alpha a + \beta b + \gamma (a \times b)$,then

Given that $\vec{a}, \vec{b}, \vec{p}$,and $\vec{q}$ are four vectors such that $\vec{a} + \vec{b} = \mu \vec{p}$,$\vec{b} \cdot \vec{q} = 0$,and $|\vec{b}|^2 = 1$,then $|(\vec{a} \cdot \vec{q}) \vec{p} - (\vec{p} \cdot \vec{q}) \vec{a}|$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module