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UPSC NDA I 2026 Mathematics Paper

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Question Numbers: 53-54

For the next two (02) items that follow:

\mathrm{P}(-2,-3,5), \mathrm{Q}(4,-1,5), \mathrm{R}(6,-4,8)

and S(2, -6, 10) are four points.

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Question : 54 of 120

Marks: +1, -0

What is

\left(\mathrm{PQ}^{2}+2 \mathrm{QS}^{2}-2 \mathrm{PR}^{2}\right)

equal to?

-1
0
1
2

Solution:

Concept:

Use 3D distance formula:

AB^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2

.

Explanation:

\mathrm{PQ}^2 = (4+2)^2 + (-1+3)^2 + (5-5)^2 = 6^2+2^2+0^2 = 40

\mathrm{QS}^2 = (2-4)^2 + (-6+1)^2 + (10-5)^2 = (-2)^2+(-5)^2+5^2 = 54

\mathrm{PR}^2 = (6+2)^2 + (-4+3)^2 + (8-5)^2 = 8^2+(-1)^2+3^2 = 74

So,

\mathrm{PQ}^2 + 2\mathrm{QS}^2 - 2\mathrm{PR}^2 = 40 + 2(54) - 2(74) = 0

Answer:

0

(Option B)

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