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The length of line segment joining the points with position vectors 2i^−2j^+3k^2\hat{i} - 2\hat{j} + 3\hat{k} and 5i^+2j^+3k^5\hat{i} + 2\hat{j} + 3\hat{k} is

Solution

✅ Correct Option: 4

The position vectors represent two points in 3D space:

Point A: 2i^−2j^+3k^2\hat{i} - 2\hat{j} + 3\hat{k} has coordinates (2,−2,3)(2, -2, 3)

Point B: 5i^+2j^+3k^5\hat{i} + 2\hat{j} + 3\hat{k} has coordinates (5,2,3)(5, 2, 3)


The difference vector from Point A to Point B:

AB⃗=(5i^+2j^+3k^)−(2i^−2j^+3k^)\vec{AB} = (5\hat{i} + 2\hat{j} + 3\hat{k}) - (2\hat{i} - 2\hat{j} + 3\hat{k})

For each component:

i^\hat{i} component: 5−2=35 - 2 = 3

j^\hat{j} component: 2−(−2)=42 - (-2) = 4

k^\hat{k} component: 3−3=03 - 3 = 0

Therefore:

AB⃗=3i^+4j^+0k^\vec{AB} = 3\hat{i} + 4\hat{j} + 0\hat{k}


The length is the magnitude of the difference vector:

Length=(3)2+(4)2+(0)2\text{Length} = \sqrt{(3)^2 + (4)^2 + (0)^2}

=9+16+0= \sqrt{9 + 16 + 0}

=25= \sqrt{25}

=5= 5


The length of the line segment is 55 units.

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