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← JEE Main 2022 · 30 Jul · Shift 2
MathematicsInteger+4 / -1Question 88 of 90

JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 88

Let OABC be a parallelogram, O be the origin, A (2, 4, −5) and C (b, 2, 3). If P (a, a, a) and Q (9 − a2^2, 3, a − 1), a ∈ ℕ, are two points such that the projection of OP→\overrightarrow{OP} on OB→\overrightarrow{OB} is 2 and OQ→\overrightarrow{OQ} makes acute angles with all three co-ordinate axes, then ∣OB→∣2+∣AC→∣2|\overrightarrow{OB}|^2+|\overrightarrow{AC}|^2 is equal to _______.

Answer and worked solution

Official answer: 118

Find a. Acute angles with all axes need 9−a2>09-a^2>0 and a−1>0a-1>0: a=2a=2, so P = (2, 2, 2).

Find b. OB→=A⃗+C⃗=(b+2,6,−2)\overrightarrow{OB}=\vec A+\vec C=(b+2,6,-2). Projection 2b+12(b+2)2+40=2\dfrac{2b+12}{\sqrt{(b+2)^2+40}}=2 gives b=1b=1.

Sum ∣(3,6,−2)∣2+∣(−1,−2,8)∣2=49+69=118|(3,6,-2)|^2+|(-1,-2,8)|^2=49+69=118.

Answer: 118