A tetrahedron has vertices P(1,2,1), Q(2,1,3), R(-1,1,2),… — JEE Main 2019 Mathematics PYQ

JEE Main 2019 · 12 Jan · Shift I

A tetrahedron has vertices

P(1,2,1)

Q(2,1,3)

R(-1,1,2)

O(0,0,0)

. The angle between the faces

OPQ

and

PQR

A.
$\cos^{-1}\left(\dfrac{7}{31}\right)$
B.
$\cos^{-1}\left(\dfrac{b}{35}\right)$
C.
$\cos^{-1}\left(\dfrac{19}{35}\right)$
D.
$\cos^{-1}\left(\dfrac{17}{31}\right)$

Show answer and solution

Correct option: C

Solution

Find normals to the two faces via cross products.

Face OPQ: vectors OP = (1,2,1), OQ = (2,1,3)
n₁ = OP × OQ = |i j k; 1 2 1; 2 1 3| = (6-1, 2-3, 1-4) = (5,-1,-3)

Face PQR: vectors PQ = Q-P = (1,-1,2), PR = R-P = (-2,-1,1)
n₂ = PQ × PR = |i j k; 1 -1 2; -2 -1 1| = (-1+2, -4-1, -1-2) = (1,-5,-3) ... let me recompute:
n₂ᵢ = (-1)(1) - (2)(-1) = -1+2 = 1
n₂ⱼ = -[(1)(1) - (2)(-2)] = -[1+4] = -5
n₂ₖ = (1)(-1) - (-1)(-2) = -1-2 = -3
So n₂ = (1,-5,-3)

Dot product: n₁·n₂ = (5)(1) + (-1)(-5) + (-3)(-3) = 5+5+9 = 19

Magnitudes: |n₁| = √(25+1+9) = √35, |n₂| = √(1+25+9) = √35

**cos θ = 19/(

[JEE Main 2019 · 12 Jan · Shift I]

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