Question

A particle starts moving from point 1, 1, 1 metre and reaches at point 4, 5, 13 then find out
a)initial position vector
b)final position vector
c)displacement vector
d)lengeth of shortest path​

Answer 1

hope it helps.........

Answer 2

Answer:

• i + j + k
• 4i + 5j + 13k
• 3i + 4j + 12k
• 13m

Explanation:

Given:

Initial Point (1,1,1)

Final Point (4,5,13)

To Find:

• Initial position vector
• Final position vector
• Displacement vector
• Length of the shortest path​

Chapter:

Vectors

Solution:
(a) The initial position vector or the point from where the vector begins will be formed using two points.

Initial Point (1,1,1)

Origin (0,0,0)

Thus, initial position vector = (1-0) x i + (1-0) x j + (1-0) x k

Initial position vector = 1i + 1j + 1k

(b) The Final position vector or the point where the vector ends and the origin point will be formed using these two points.

Origin (0,0,0)

Final Point (4,5,13)

Thus, final position vector = (4-0) x i + (5-0) x j + (13-0) x k

Final position vector = 4i + 5j + 13k

(c) The displacement vector is formed from the point where the final vector ends and the initial point will be formed using these two points.

Initial Point (1,1,1)

Final Point (4,5,13)

Thus, displacement vector = (4-1) x i + (5-1) x j + (13-1) x k

Displacement vector  = 3i +4j + 12k

(d) Length of the shortest path is calculated as,

Formula = $\sqrt{x^{2}+y^{2}+z^{2} }$

For finding the shortest length,

We need to square each component of the displacement vector.

Displacement vector  = 3i +4j + 12k

$x^{2} = (3i)^{2} = 9*1 = 9$

$y^{2} = (4j)^{2} = 16*1 = 16$

$z^{2} = (12k)^{2} = 144*1 = 144$

Putting in the formula,

$\sqrt{x^{2} +y^{2}+z^{2} }=\sqrt{9+16+144}$

$=\sqrt{169}$

Length of shortest path​ = 13m

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