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JEE Main 2019 · 10 April · Shift I

JEE Main 2019 Mathematics question on Vector Algebra

Let A(3,0,−1)A(3, 0, -1), B(2,10,6)B(2, 10, 6) and C(1,2,1)C(1, 2, 1) be the vertices of a triangle and MM be the mid-point of ACAC. If GG divides BMBM in the ratio 2:12:1, then cos⁡(∠GOA)\cos(\angle GOA) (OO being the origin) is equal to
  1. A.
    115\dfrac{1}{\sqrt{15}}
  2. B.
    1215\dfrac{1}{2\sqrt{15}}
  3. C.
    130\dfrac{1}{\sqrt{30}}
  4. D.
    1610\dfrac{1}{6\sqrt{10}}
Show answer and solution

Correct option: A

Solution

M is midpoint of AC: M = ((3+1)/2, (0+2)/2, (-1+1)/2) = (2, 1, 0).

G divides BM in ratio 2:1, so G = (1·B + 2·M... wait, G = (2·M + 1·B)/3 = ((4+2)/3, (2+10)/3, (0+6)/3) = (2, 4, 2).

Actually G is the centroid of triangle ABC: G = ((3+2+1)/3, (0+10+2)/3, (-1+6+1)/3) = (2, 4, 2).

Now OG = (2, 4, 2), OA = (3, 0, -1).

cos(∠GOA) = (OG · OA)/(|OG||OA|) = (6 + 0 - 2)/(√24 · √10) = 4/(√240) = 4/(4√15) = 1/√15.

[JEE Main 2019 · 10 April · Shift I]

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