### Video Transcript

A particle moves such that its
position vector π« is equal to four π‘ minus two π’ plus three π‘ plus one π£. The magnitude of its displacement
till π‘ equals two equals what length units. Is it (A) seven, (B) 10, (C) 13, or
(D) root 85?

In this question, weβre given the
position vector of a particle, and we need to calculate the magnitude of its
displacement at π‘ equals two. We recall that the displacement
vector π¬ of π‘ is equal to π« of π‘ minus π« of zero, where π« of zero is the
initial position of the particle. As we need to find the displacement
when π‘ equals two, π¬ of two will be equal to π« of two minus π« of zero. We can calculate π« of two and π«
of zero by substituting two and zero into our equation for the position vector. π« of two is equal to four
multiplied by two minus two π’ plus three multiplied by two plus one π£. This simplifies to six π’ plus
seven π£.

π« of zero is equal to four
multiplied by zero minus two π’ plus three multiplied by zero plus one π£, where π’
and π£ are the standard perpendicular unit vectors. This simplifies to negative two π’
plus π£. The displacement at π‘ equals two
is therefore equal to six π’ plus seven π£ minus negative two π’ plus π£. Six π’ minus negative two π’ is
equal to eight π’, and seven π£ minus π£ is equal to six π£. π¬ of two is equal to eight π’ plus
six π£.

We are asked to calculate the
magnitude of this. And we know that the magnitude of
any vector is equal to the square root of the sum of the squares of its individual
components. The magnitude of the displacement
is equal to the square root of eight squared plus six squared. Eight squared is equal to 64, and
six squared is 36. So we have the square root of
100. This is equal to 10. We can therefore conclude that the
magnitude of the displacement of the particle till π‘ equals two is equal to 10
length units. The correct answer is option
(B).