If a, b and c are three vectors such that |a | vedclass

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

Question

(D) Given $|a+b+c|^2 = 1^2 = 1$. Since $c = \lambda(a 	imes b)$, $c$ is perpendicular to both $a$ and $b$, so $a \cdot c = 0$ and $b \cdot c = 0$.Exp

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) Given $|a+b+c|^2 = 1^2 = 1$. Since $c = \lambda(a 	imes b)$, $c$ is perpendicular to both $a$ and $b$, so $a \cdot c = 0$ and $b \cdot c = 0$.Expanding $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 1$.Substituting the values: $(\frac{1}{\sqrt{3}})^2 + (\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{6}})^2 + 2(a \cdot b + 0 + 0) = 1$.$\frac{1}{3} + \frac{1}{2} + \frac{1}{6} + 2(a \cdot b) = 1$.$\frac{2+3+1}{6} + 2(a \cdot b) = 1 \implies 1 + 2(a \cdot b) = 1 \implies a \cdot b = 0$.Since $a \cdot b = |a||b| \cos 	heta = 0$ and $|a|, |b| 
eq 0$, we have $\cos 	heta = 0$, so $	heta = \frac{\pi}{2}$.

If $a, b$ and $c$ are three vectors such that $|a + b + c| = 1$, $c = \lambda(a \times b)$, and $|a| = \frac{1}{\sqrt{3}}$, $|b| = \frac{1}{\sqrt{2}}$, $|c| = \frac{1}{\sqrt{6}}$, then the angle between $a$ and $b$ is

Similar Questions

If $|a| = |b| = 1$ and $|a + b| = \sqrt{3}$,then the value of $(3a - 4b) \cdot (2a + 5b)$ is

The vectors $(a \cdot b) c$ and $(a \cdot c) b$ are:

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=\sqrt{14}$,$|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^2$ is equal to $...........$.

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $|\vec{a}+3 \vec{b}-4 \vec{c}|^2=$

Using vectors,prove that $\cos (A-B)=\cos A \cos B+\sin A \sin B$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module