CUET UG 2025 — Mathematics Algebra

The position vectors represent two points in 3D space:

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

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

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The difference vector from Point A to Point B:

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

For each component:

$\hat{i}$i^ component: $5 - 2 = 3$5−2=3

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

$\hat{k}$k^ component: $3 - 3 = 0$3−3=0

Therefore:

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

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The length is the magnitude of the difference vector:

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

$$= \sqrt{9 + 16 + 0}$$=9+16+0​

$$= \sqrt{25}$$=25​

$$= 5$$=5

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The length of the line segment is $5$5 units.