A particle undergoes a displacement r -2-0 -3-0 -6-0 k- ending with the position vector r -3-0 -4-0 k- in meters- What was its initial position vector-A- 3 -4 k-B- 2 -2 k-C- 2 -6 -10 k-D- -2 -6 -10 k-

The correct option is D r=r_f r_i r_i=r_f r =3ĵ 4k̂ (2î 3ĵ+6k̂) r_i= 2î+6ĵ 10k̂ So the particle started from the initial position whose position vector is g ...

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Byju's Answer

Standard XII

Physics

Displacement

A particle un...

Question

A particle undergoes a displacement $△ \to r = 2.0^i - 3.0^j + 6.0^k$ , ending with the position vector $\to r = 3.0^j - 4.0^k$ , in meters, What was its initial position vector?

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Solution

The correct option is D

$△ r = r_{f} - r_{i}$
$r_{i} = r_{f} - △ r$
$= 3^j - 4^k - (2^i - 3^j + 6^k)$
$r_{i} = - 2^i + 6^j - 10^k$

So the particle started from the initial position whose position vector is given by
$r_{i} = - 2^i + 6^j - 10^k$

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Similar questions

A particle undergoes a displacement $△ \to r = 2.0^i - 3.0^j + 6.0^k$ , ending with the position vector $\to r = 3.0^j - 4.0^k$ , in meters, What was its initial position vector?

Q. The position vector of four identical masses of mass

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_{1} = (^i + 2^j + 7^k),_{2} = (3^i + 5^j +^k),

_{3} = (6^i + 2^j + 3^k)

and

_{4} = (2^i -^j + 5^k)

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Two particles of mass 1 kg and 3 kg have position vectors $2^i + 3^j + 4^k a n d - 2^i + 3^j - 4^k$ respectively. The centre of mass has a position vector

Q. Find the value of

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3^i + 6^j + 9^k,^i + 2^j + 3^k, 2^i + 3^j +^k

and

4^i + 6^j + λ^k

are coplanar.

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A particle undergoes a displacement △→r=2.0^i−3.0^j+6.0^k, ending with the position vector →r=3.0^j−4.0^k, in meters, What was its initial position vector?

The correct option is D △r=rf−ri ri=rf−△r =3^j−4^k−(2^i−3^j+6^k) ri=−2^i+6^j−10^k So the particle started from the initial position whose position vector is given by ri=−2^i+6^j−10^k

A particle undergoes a displacement $△ \to r = 2.0^i - 3.0^j + 6.0^k$ , ending with the position vector $\to r = 3.0^j - 4.0^k$ , in meters, What was its initial position vector?

The correct option is D

$△ r = r_{f} - r_{i}$
$r_{i} = r_{f} - △ r$
$= 3^j - 4^k - (2^i - 3^j + 6^k)$
$r_{i} = - 2^i + 6^j - 10^k$

So the particle started from the initial position whose position vector is given by
$r_{i} = - 2^i + 6^j - 10^k$