Let x and y be real numbers. If vec{a}=(sin x | vedclass

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

Question

(C) Given,$\vec{a}=(\sin x) \hat{i}+(\sin y) \hat{j}$ and $\vec{b}=(\cos x) \hat{i}+(\cos y) \hat{j}$.We calculate the cross product $\vec{a} 	imes \

Vedclass · Accepted Answer

Less than or equal to $1$

Vedclass · Answer

(C) Given,$\vec{a}=(\sin x) \hat{i}+(\sin y) \hat{j}$ and $\vec{b}=(\cos x) \hat{i}+(\cos y) \hat{j}$.We calculate the cross product $\vec{a} 	imes \vec{b}$ as follows:$\vec{a} 	imes \vec{b} = ((\sin x) \hat{i}+(\sin y) \hat{j}) 	imes ((\cos x) \hat{i}+(\cos y) \hat{j})$$= (\sin x \cos y) (\hat{i} 	imes \hat{i}) + (\sin x \cos y) (\hat{i} 	imes \hat{j}) + (\sin y \cos x) (\hat{j} 	imes \hat{i}) + (\sin y \cos y) (\hat{j} 	imes \hat{j})$Since $\hat{i} 	imes \hat{i} = 0$,$\hat{j} 	imes \hat{j} = 0$,$\hat{i} 	imes \hat{j} = \hat{k}$,and $\hat{j} 	imes \hat{i} = -\hat{k}$:$\vec{a} 	imes \vec{b} = (\sin x \cos y) \hat{k} - (\sin y \cos x) \hat{k} = (\sin x \cos y - \cos x \sin y) \hat{k} = \sin(x-y) \hat{k}$.Now,the magnitude is $|\vec{a} 	imes \vec{b}| = |\sin(x-y)|$.Since the range of the sine function is $[-1, 1]$,the absolute value $|\sin(x-y)|$ must be in the interval $[0, 1]$.Therefore,$|\vec{a} 	imes \vec{b}| \leq 1$.

Let $x$ and $y$ be real numbers. If $\vec{a}=(\sin x) \hat{i}+(\sin y) \hat{j}$ and $\vec{b}=(\cos x) \hat{i}+(\cos y) \hat{j}$,then $|\vec{a} \times \vec{b}|$ is

Similar Questions

If $|a|=1, |b|=2$ and the angle between $a$ and $b$ is $120^{\circ}$, then ${(a+3b) \times (3a-b)}^2$ is equal to

If $1, 2, 3$ and $-1, 0, 1$ are the direction ratios of the rays $OA$ and $OB$ respectively, then the direction cosines of a normal to the plane $AOB$ are

The unit vectors perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ are:

Let $L_1: \frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}$ and $L_2: \frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3}$ be the given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

The area of the parallelogram whose diagonals are $\frac{3}{2}i + \frac{1}{2}j - k$ and $2i - 6j + 8k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module