Question

A body moves 6m north and 8m east and 10m vertically upwards,the resultant displacement of the body from its initial position is:

$1) \: 10 \sqrt{2} \: m \\ 2)10 \: m \\ 3) \frac{10}{ \sqrt{2} } m \\ 4)20 \: m$

Explanation Required

Thank Yuh​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• It moves "6m" North .
• It moves "8m" East.
• It moves "10m" vertically upward.

$\huge{\bold{\underline{Explanation:-}}}$

# refer the attachment for figure.

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This is the case of 3 - D motion,

I.e. Motion in a plane,

Here we apply the formula of Position vector.

From the formula,

Magnitude of displacement =

$\large{\boxed{\tt r = \sqrt{{r_x}^2 + {r_y}^2 + {r_z}^2}}}$

Here:-

$\large{\tt r = displacement \: of \: the \: particle}$

Now,

• $\large{\tt r_x = 8 \: m}$
• $\large{\tt r_y = 6 \: m}$
• $\large{\tt r_z = 10 \: m}$

[Here I have taken the measurements according to the diagram I have prepared]

Substituting the values in the formula,

$\large{\tt r = \sqrt{(8)^2 + (6)^2 + (10)^2}}$

Now,

$\large{\tt r = \sqrt{64 + 36 + 100}}$

$\large{\tt r = \sqrt{100 + 100}}$

$\large{\tt r = \sqrt{200}}$

$\huge{\boxed{\boxed{\tt r = 10 \sqrt{2} \: meters}}}$

So, the resultant displacement is 10 √2 meters (Option - 1).

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Additional information:-

• Position vector:-

To locate any particle in any plane , we use a coordinate system,

The location of the particle through the coordinate system is called Position vector.

For example,

$\large{\vec{A} = x\hat{i} + y \hat{j}}$

• Co - planar vectors:-

The vectors lying in the same plane is called co - planar vectors.

• Orthogonal vectors:-

Two vectors are said to be orthogonal if the angle between two vectors is 90°.

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Answer 2

Answer:

Option => 1.

$\sf 10 \sqrt{2}\;m$

Explanation:

Given :

A body moves :

• 6m north.
• 8m east.
• 10m.

A body moves vertically upwards.

To Find :

The resultant displacement of the body from its initial position is.

Solution :

• For magnitude as well as direction of displacement.

Now,

Consider the :

To represent unit vectors in x,y,z axis as - i , j , k

As given,

Initially the body was at origin.

So,

It's position vector is R1 = $\bf 0i+0j+0k.$

Now,

After the motion Position is R2 = $\bf 8i+6j+10k.$

Hence,

Change in position vector becomes:

$\sf \implies R=R2-R1$

Now,

$\sf \implies R=8i+6j+10k$

We know :

Magnitude of displacement vector :

The distance from the starting point to the ending point is the magnitude of the displacement vector.

So,

$\bigstar\;\boxed{\sf \sqrt{(8^2+6^2+10^2)}}}$

Now,

$\sf\\ \\\implies \sqrt{(8^2+6^2+10^2)}\\ \\ \\\implies \sqrt{64+12+100}\\ \\ \\ \implies \sqrt{100+100}}\\ \\ \\ \implies \sqrt{200}\\ \\ \\ \implies 10\sqrt{2}\;m$

Hence,

The Option 1 is correct.

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