Question:
Two vectors are given by $\vec{A}=\left(3\hat{i} +\hat{j}+3\hat{k}\right)$ and $ \vec{B}=\left(3\hat{i} +5\hat{j}-2\hat{k}\right)$. Find the third vector $\vec{C}$ if $\vec{A}+3\vec{B}-\vec{C}=0.$
Two vectors are given by $\vec{A}=\left(3\hat{i} +\hat{j}+3\hat{k}\right)$ and $ \vec{B}=\left(3\hat{i} +5\hat{j}-2\hat{k}\right)$. Find the third vector $\vec{C}$ if $\vec{A}+3\vec{B}-\vec{C}=0.$
Updated On: Jun 6, 2024
- $\left(12 \hat{i}+14 \hat{j}+12 \hat{k}\right)$
- $\left(13 \hat{i}+17 \hat{j}+12 \hat{k}\right)$
- $\left(12 \hat{i}+16 \hat{j}-3 \hat{k}\right)$
- $\left(15 \hat{i}+13 \hat{j}+4 \hat{k}\right)$
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The Correct Option is C
Solution and Explanation
$\vec{A}+3\vec{B}-\vec{C}=0$
$3\hat{i}+\hat{j}+3\hat{k}+3\left(3 \hat{i}+5\hat{j}-2 \hat{k}\right)-\vec{C}=0$
$12 \hat{i}+16 \hat{j}-3 \hat{k}-\vec{C}=0$,
$ \vec{C}=12 \hat{i}+16 \hat{j}-3 \hat{k}$
$3\hat{i}+\hat{j}+3\hat{k}+3\left(3 \hat{i}+5\hat{j}-2 \hat{k}\right)-\vec{C}=0$
$12 \hat{i}+16 \hat{j}-3 \hat{k}-\vec{C}=0$,
$ \vec{C}=12 \hat{i}+16 \hat{j}-3 \hat{k}$
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration