Let vec a, vec b, vec c be three vectors such | vedclass

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

Question

(C) Given that $\vec a \perp (\vec b + \vec c)$,$\vec b \perp (\vec c + \vec a)$,and $\vec c \perp (\vec a + \vec b)$.This implies the dot products ar

Vedclass · Accepted Answer

$\sqrt{14}$

Vedclass · Answer

(C) Given that $\vec a \perp (\vec b + \vec c)$,$\vec b \perp (\vec c + \vec a)$,and $\vec c \perp (\vec a + \vec b)$.This implies the dot products are zero:$\vec a \cdot (\vec b + \vec c) = 0 \Rightarrow \vec a \cdot \vec b + \vec a \cdot \vec c = 0$$\vec b \cdot (\vec c + \vec a) = 0 \Rightarrow \vec b \cdot \vec c + \vec b \cdot \vec a = 0$$\vec c \cdot (\vec a + \vec b) = 0 \Rightarrow \vec c \cdot \vec a + \vec c \cdot \vec b = 0$Adding these equations,we get $2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a) = 0$,which implies $\vec a \cdot \vec b = \vec b \cdot \vec c = \vec c \cdot \vec a = 0$.Now,consider $|\vec a + \vec b + \vec c|^2 = |\vec a|^2 + |\vec b|^2 + |\vec c|^2 + 2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a)$.Substituting the given magnitudes and the dot products:$|\vec a + \vec b + \vec c|^2 = 1^2 + 2^2 + 3^2 + 2(0) = 1 + 4 + 9 = 14$.Therefore,$|\vec a + \vec b + \vec c| = \sqrt{14}$.

Let $\vec a, \vec b, \vec c$ be three vectors such that $\vec a \perp (\vec b + \vec c)$,$\vec b \perp (\vec c + \vec a)$,and $\vec c \perp (\vec a + \vec b)$. If $|\vec a| = 1, |\vec b| = 2, |\vec c| = 3$,then $|\vec a + \vec b + \vec c| = \dots$

Similar Questions

In a trapezium,the vector $\overrightarrow{BC} = \lambda \overrightarrow{AD}$. We will then find that $p = \overrightarrow{AC} + \overrightarrow{BD}$ is collinear with $\overrightarrow{AD}$. If $p = \mu \overrightarrow{AD}$,then

If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$,$|\vec{a}| = 3$,$|\vec{b}| = 5$,and $|\vec{c}| = 7$,find the angle between $\vec{a}$ and $\vec{b}$.

Let $u, v, w$ be three vectors such that $|u| = 1, |v| = 2, |w| = 3$. If the projection of $v$ along $u$ is equal to the projection of $w$ along $u$ and $v, w$ are perpendicular to each other, then $|u - v + w| = ...$

Let $a = i + 2j + k$, $b = i - j + k$, and $c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has a projection of $\frac{1}{\sqrt{3}}$ on $c$. Then, one such vector is:

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module