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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2026 · Mathematics · Three Dimensional Geometry

Passage: P(-2,-3,5) , Q(4,-1,5) , R(6,-4,8) and S(2,-6,10) are four...

Passage: P(−2,−3,5)P(-2,-3,5), Q(4,−1,5)Q(4,-1,5), R(6,−4,8)R(6,-4,8) and S(2,−6,10)S(2,-6,10) are four points.

Question: What is (PQ2+2QS2−2PR2)(PQ^2+2QS^2-2PR^2) equal to ?

Options

  1. A−1-1
  2. B00
  3. C11
  4. D22

Correct answer

B. 00

Step-by-step solution

Given points are P(−2,−3,5)P(-2,-3,5), Q(4,−1,5)Q(4,-1,5), R(6,−4,8)R(6,-4,8) and S(2,−6,10)S(2,-6,10).

Using the distance formula, we find the squares of the distances between the required points:

PQ2=(4−(−2))2+(−1−(−3))2+(5−5)2=62+22+02=36+4=40PQ^2 = (4 - (-2))^2 + (-1 - (-3))^2 + (5 - 5)^2 = 6^2 + 2^2 + 0^2 = 36 + 4 = 40

QS2=(2−4)2+(−6−(−1))2+(10−5)2=(−2)2+(−5)2+52=4+25+25=54QS^2 = (2 - 4)^2 + (-6 - (-1))^2 + (10 - 5)^2 = (-2)^2 + (-5)^2 + 5^2 = 4 + 25 + 25 = 54

PR2=(6−(−2))2+(−4−(−3))2+(8−5)2=82+(−1)2+32=64+1+9=74PR^2 = (6 - (-2))^2 + (-4 - (-3))^2 + (8 - 5)^2 = 8^2 + (-1)^2 + 3^2 = 64 + 1 + 9 = 74

Substituting these values into the given expression:

PQ2+2QS2−2PR2=40+2(54)−2(74)PQ^2 + 2QS^2 - 2PR^2 = 40 + 2(54) - 2(74)

=40+108−148=0= 40 + 108 - 148 = 0

Answer: 00
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