Given vectors vec{a} = 2 hat{i} - hat{j} + 2 | vedclass

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

Question

(A) First,calculate the sum of the vectors: $\vec{a} + \vec{b} = (2 \hat{i} - \hat{j} + 2 \hat{k}) + (-\hat{i} + \hat{j} - \hat{k}) = (2-1) \hat{i} +

Vedclass · Accepted Answer

$\hat{i} + \hat{k}$

Vedclass · Answer

(A) First,calculate the sum of the vectors: $\vec{a} + \vec{b} = (2 \hat{i} - \hat{j} + 2 \hat{k}) + (-\hat{i} + \hat{j} - \hat{k}) = (2-1) \hat{i} + (-1+1) \hat{j} + (2-1) \hat{k} = \hat{i} + \hat{k}$.Next,find the magnitude of the resultant vector: $|\vec{a} + \vec{b}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$.The unit vector in the direction of $\vec{a} + \vec{b}$ is $\hat{u} = \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} = \frac{\hat{i} + \hat{k}}{\sqrt{2}}$.The required vector with magnitude $\sqrt{2}$ is $\sqrt{2} 	imes \hat{u} = \sqrt{2} 	imes \frac{\hat{i} + \hat{k}}{\sqrt{2}} = \hat{i} + \hat{k}$.Thus,the correct option is $A$.

Given vectors $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} - \hat{k}$. The vector in the direction of $\vec{a} + \vec{b}$ with magnitude $\sqrt{2}$ is . . . . . . .

Similar Questions

Let $p = (x + 4y)\vec{a} + (2x + y + 1)\vec{b}$ and $q = (y - 2x + 2)\vec{a} + (2x - 3y - 1)\vec{b}$,where $\vec{a}$ and $\vec{b}$ are non-collinear vectors. If $3p = 2q$,then the values of $x$ and $y$ are:

Three non-zero non-collinear vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are such that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$, and $3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$. Then $\vec{a}+3\vec{b}+2\vec{c}$ equals to

Consider the following statements :
Statement $(I)$ : If either $|\vec{a}|=0$ or $|\vec{b}|=0$,then $\vec{a} \cdot \vec{b}=0$.
Statement $(II)$ : If $\vec{a} \times \vec{b}=\vec{0}$,then $\vec{a}$ is perpendicular to $\vec{b}$.
Which of the following is correct?

Write two different vectors having the same magnitude.

The position vectors of $A$ and $B$ are $(\hat{i}+\hat{j}+\hat{k})$ and $(\frac{1}{3} \hat{j}+\frac{1}{3} \hat{k})$. If $B$ divides the line segment $AC$ in the ratio $2:1$,then the position vector of $C$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module