Magnitude of a given vector with end points (4, -4, 0) and (-2, -2, 0) must be

English

Gujarati

Hindi

3-1.Vectors

easy

The magnitude of a given vector with end points $ (4, -4, 0)$ and $(-2, -2, 0)$ must be

A

$6$

B

$5\sqrt 2 $

C

$4$

D

$2\sqrt {10} $

Solution

(d)$\vec r = {\vec r_2} – {\vec r_1} = ( – 2\hat i – 2\hat j + 0\hat k) – (4\hat i – 4\hat j + 0\hat k)$

$⇒$ $\vec r = – 6\hat i + 2\hat j + 0\hat k$

$\therefore |\vec r| = \sqrt {{{( – 6)}^2} + {{(2)}^2} + {0^2}} = \sqrt {36 + 4} = \sqrt {40} = 2\sqrt {10} $

Standard 11

Physics

Share

Similar Questions

If $\overrightarrow A = 4\hat i – 3\hat j$ and $\overrightarrow B = 6\hat i + 8\hat j$ then magnitude and direction of $\overrightarrow A \, + \overrightarrow B $ will be

medium

Two vectors $\vec A$ and $\vec B$ have equal magnitudes. The magnitude of $(\vec A + \vec B)$ is $‘n’$ times the magnitude of $(\vec A – \vec B)$. The angle between $ \vec A$ and $\vec B$ is

hard

(JEE MAIN-2021)

A vector $\vec A $ is rotated by a small angle $\Delta \theta$ radian $( \Delta \theta << 1)$ to get a new vector $\vec B$ In that case $\left| {\vec B – \vec A} \right|$ is

medium

(JEE MAIN-2015)

If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $\mathrm{R}$ are inclined at an angle $\theta$, then

medium

(JEE MAIN-2024)

The resultant force of $5 \,N$ and $10 \,N$ can not be …….. $N$

easy