Question:
The $(x, y, z)$ coordinates of two points $A$ and $B$ are given respectively as $(0, 3,-1)$ and $(-2, 6, 4)$. The displacement vector form $A$ to $B$ may be given by:
The $(x, y, z)$ coordinates of two points $A$ and $B$ are given respectively as $(0, 3,-1)$ and $(-2, 6, 4)$. The displacement vector form $A$ to $B$ may be given by:
Updated On: Aug 25, 2023
- $-2 \hat{i}+6 j+4 \hat{k}$
- $-2 \hat{i}+3 j+3 \hat{k}$
- $-2 \hat{i}+3 j+5 \hat{k}$
- $2 \hat{i} - 3 j - 5 \hat{k}$
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The Correct Option is C
Solution and Explanation
The correct option is(C): \(=-2 \hat{ i }+3 \hat{ j }+5 \hat{ k }.\)
The displacement vector from
\(A \equiv(0,3,-1) \text { to } B \equiv(-2,6,4)\) is
\(=(-2 \hat{ i }+6 \hat{ j }+4 \hat{ k })-(3 \hat{ j }-\hat{ k })\)
\(=-2 \hat{ i }+3 \hat{ j }+5 \hat{ k }\)
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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