Find the unit vector in the direction of the | vedclass

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

Question

(A) Let $\vec{c} = \vec{a} + \vec{b}$.Given $\vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$.Then $\vec{c} = (

Vedclass · Accepted Answer

$\frac{4}{\sqrt{29}} \hat{i}+\frac{3}{\sqrt{29}} \hat{j}-\frac{2}{\sqrt{29}} \hat{k}$

Vedclass · Answer

(A) Let $\vec{c} = \vec{a} + \vec{b}$.Given $\vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$.Then $\vec{c} = (2+2)\hat{i} + (2+1)\hat{j} + (-5+3)\hat{k} = 4\hat{i} + 3\hat{j} - 2\hat{k}$.The magnitude of $\vec{c}$ is $|\vec{c}| = \sqrt{4^2 + 3^2 + (-2)^2} = \sqrt{16 + 9 + 4} = \sqrt{29}$.The unit vector in the direction of $\vec{c}$ is given by $\hat{c} = \frac{\vec{c}}{|\vec{c}|}$.$\hat{c} = \frac{1}{\sqrt{29}}(4\hat{i} + 3\hat{j} - 2\hat{k}) = \frac{4}{\sqrt{29}}\hat{i} + \frac{3}{\sqrt{29}}\hat{j} - \frac{2}{\sqrt{29}}\hat{k}$.

Find the unit vector in the direction of the sum of the vectors,$\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+3 \hat{k}$.

Similar Questions

Let $G$ be the centroid of a triangle $ABC$ and $O$ be any other point in that plane,then $\overline{OA}+\overline{OB}+\overline{OC}+\overline{OG}=$

In a quadrilateral $ABCD$, the point $P$ divides $DC$ in the ratio $1:3$ internally and $Q$ is the mid-point of $AC$. If $\vec{AB} + \vec{AD} + \vec{BC} - 2\vec{DC} = \lambda \vec{PQ}$, then the value of $\lambda$ is

Let $\vec{\alpha} = (\lambda - 2) \vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$ be two given vectors where $\vec{a}$ and $\vec{b}$ are non-collinear. The value of $\lambda$ for which vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear is:

$P$ and $Q$ are the points of trisection of the line segment $AB$. If $2 \hat{i}-5 \hat{j}+3 \hat{k}$ and $4 \hat{i}+\hat{j}-6 \hat{k}$ are the position vectors of $A$ and $B$ respectively, then the position vector of the point which divides $PQ$ in the ratio $2:3$ is

If $PQRST$ is a pentagon,then the resultant of forces $\overline{PQ}, \overline{PT}, \overline{QR}, \overline{SR}, \overline{TS}$ and $\overline{PS}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module