Question Video: Finding the Magnitude of the Displacement Vector | Nagwa Question Video: Finding the Magnitude of the Displacement Vector | Nagwa

# Question Video: Finding the Magnitude of the Displacement Vector Mathematics • Second Year of Secondary School

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A particle moves such that its position vector π« = (4π‘ β 2)π’ + (3π‘ + 1)π£. The magnitude of its displacement till π‘ = 2 equals οΌΏ length units. [A] 7 [B] 10 [C] 13 [D] β85

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### 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, weβre 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).

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