Let hat{u} and hat{v} be unit vectors incline | vedclass

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

Question

(A) Given $\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} 	imes \hat{v})$.Since $\hat{u}$ and $\hat{v}$ are unit vectors,$|\hat{u} 	imes \hat{v}| =

Vedclass · Accepted Answer

$\frac{4}{3}(\vec{A} \cdot \hat{u}) - \frac{2}{3}(\vec{A} \cdot \hat{v})$

Vedclass · Answer

(A) Given $\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} 	imes \hat{v})$.Since $\hat{u}$ and $\hat{v}$ are unit vectors,$|\hat{u} 	imes \hat{v}| = |\hat{u}||\hat{v}| \sin 	heta = \sin 	heta = \frac{\sqrt{3}}{2}$.Since $	heta$ is acute,$	heta = 60^\circ = \frac{\pi}{3}$.Thus,$\hat{u} \cdot \hat{v} = \cos 60^\circ = \frac{1}{2}$.Taking the dot product of $\vec{A}$ with $\hat{u}$: $\vec{A} \cdot \hat{u} = \lambda(\hat{u} \cdot \hat{u}) + (\hat{v} \cdot \hat{u}) + ((\hat{u} 	imes \hat{v}) \cdot \hat{u}) = \lambda + \frac{1}{2} + 0 = \lambda + \frac{1}{2}$.Taking the dot product of $\vec{A}$ with $\hat{v}$: $\vec{A} \cdot \hat{v} = \lambda(\hat{u} \cdot \hat{v}) + (\hat{v} \cdot \hat{v}) + ((\hat{u} 	imes \hat{v}) \cdot \hat{v}) = \frac{\lambda}{2} + 1 + 0 = \frac{\lambda}{2} + 1$.Now,evaluate option $A$: $\frac{4}{3}(\vec{A} \cdot \hat{u}) - \frac{2}{3}(\vec{A} \cdot \hat{v}) = \frac{4}{3}(\lambda + \frac{1}{2}) - \frac{2}{3}(\frac{\lambda}{2} + 1) = \frac{4\lambda + 2 - \lambda - 2}{3} = \frac{3\lambda}{3} = \lambda$.

Let $\hat{u}$ and $\hat{v}$ be unit vectors inclined at an acute angle such that $|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}$. If $\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v})$,then $\lambda$ is equal to:

Similar Questions

If $p$-th,$q$-th,and $r$-th terms of a geometric progression are the positive numbers $a, b,$ and $c$ respectively,then the angle between the vectors $(\log a^2) i + (\log b^2) j + (\log c^2) k$ and $(q-r) i + (r-p) j + (p-q) k$ is

If $\bar{a}, \bar{b}$ and $\bar{c}$ are vectors such that $|\bar{a}| = |\frac{\bar{b}}{2}| = |\frac{\bar{c}}{3}| = 1$; $\bar{b}$ and $\bar{c}$ are perpendicular; and the projections of $\bar{b}$ and $\bar{c}$ on $\bar{a}$ are equal,then $|\bar{a} - \bar{b} + \bar{c}| = $

The shortest distance between the lines $r = 3i + 5j + 7k + \lambda(i + 2j + k)$ and $r = -i - j - k + \mu(7i - 6j + k)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module