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Question: In a tetrahedron OABC, the edges are of lengths, |OA| = |BC| = a, |OB| = |AC| = b, |OC| = |AB| = c. ...

In a tetrahedron OABC, the edges are of lengths, |OA| = |BC| = a, |OB| = |AC| = b, |OC| = |AB| = c. Let G₁ and G₂ be the centroids of the triangle ABC and AOC such that OG1⊥BG2OG_1 \perp BG_2, then the value of a2+c2b2\frac{a^2+c^2}{b^2} is (O being origin)

A

2

B

3

C

6

D

9

Answer

3

Explanation

Solution

The position vectors of the vertices of the tetrahedron OABC, with O as the origin, are 0⃗\vec{0}, a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c} respectively.

The lengths of the edges are given as:

∣OA∣=∣a⃗∣=a|OA| = |\vec{a}| = a

∣OB∣=∣b⃗∣=b|OB| = |\vec{b}| = b

∣OC∣=∣c⃗∣=c|OC| = |\vec{c}| = c

∣BC∣=∣c⃗−b⃗∣=a|BC| = |\vec{c} - \vec{b}| = a

∣AC∣=∣c⃗−a⃗∣=b|AC| = |\vec{c} - \vec{a}| = b

∣AB∣=∣b⃗−a⃗∣=c|AB| = |\vec{b} - \vec{a}| = c

From the squared magnitudes of the difference vectors, we can find the dot products of the position vectors:

∣c⃗−b⃗∣2=a2  ⟹  (c⃗−b⃗)⋅(c⃗−b⃗)=a2  ⟹  ∣c⃗∣2−2b⃗⋅c⃗+∣b⃗∣2=a2  ⟹  c2−2b⃗⋅c⃗+b2=a2|\vec{c} - \vec{b}|^2 = a^2 \implies (\vec{c} - \vec{b}) \cdot (\vec{c} - \vec{b}) = a^2 \implies |\vec{c}|^2 - 2\vec{b} \cdot \vec{c} + |\vec{b}|^2 = a^2 \implies c^2 - 2\vec{b} \cdot \vec{c} + b^2 = a^2

Thus, 2b⃗⋅c⃗=b2+c2−a22\vec{b} \cdot \vec{c} = b^2 + c^2 - a^2.

∣c⃗−a⃗∣2=b2  ⟹  (c⃗−a⃗)⋅(c⃗−a⃗)=b2  ⟹  ∣c⃗∣2−2a⃗⋅c⃗+∣a⃗∣2=b2  ⟹  c2−2a⃗⋅c⃗+a2=b2|\vec{c} - \vec{a}|^2 = b^2 \implies (\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = b^2 \implies |\vec{c}|^2 - 2\vec{a} \cdot \vec{c} + |\vec{a}|^2 = b^2 \implies c^2 - 2\vec{a} \cdot \vec{c} + a^2 = b^2

Thus, 2a⃗⋅c⃗=a2+c2−b22\vec{a} \cdot \vec{c} = a^2 + c^2 - b^2.

∣b⃗−a⃗∣2=c2  ⟹  (b⃗−a⃗)⋅(b⃗−a⃗)=c2  ⟹  ∣b⃗∣2−2a⃗⋅b⃗+∣a⃗∣2=c2  ⟹  b2−2a⃗⋅b⃗+a2=c2|\vec{b} - \vec{a}|^2 = c^2 \implies (\vec{b} - \vec{a}) \cdot (\vec{b} - \vec{a}) = c^2 \implies |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{a}|^2 = c^2 \implies b^2 - 2\vec{a} \cdot \vec{b} + a^2 = c^2

Thus, 2a⃗⋅b⃗=a2+b2−c22\vec{a} \cdot \vec{b} = a^2 + b^2 - c^2.

G1G_1 is the centroid of triangle ABC. Its position vector is g1⃗=a⃗+b⃗+c⃗3\vec{g_1} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}.

G2G_2 is the centroid of triangle AOC. Its position vector is g2⃗=0⃗+a⃗+c⃗3=a⃗+c⃗3\vec{g_2} = \frac{\vec{0} + \vec{a} + \vec{c}}{3} = \frac{\vec{a} + \vec{c}}{3}.

We are given that OG1⊥BG2OG_1 \perp BG_2. The vector OG1OG_1 is g1⃗−0⃗=g1⃗\vec{g_1} - \vec{0} = \vec{g_1}. The vector BG2BG_2 is g2⃗−b⃗\vec{g_2} - \vec{b}.

The condition of perpendicularity is g1⃗⋅(g2⃗−b⃗)=0\vec{g_1} \cdot (\vec{g_2} - \vec{b}) = 0.

Substitute the expressions for g1⃗\vec{g_1} and g2⃗\vec{g_2}:

(a⃗+b⃗+c⃗3)⋅(a⃗+c⃗3−b⃗)=0\left(\frac{\vec{a} + \vec{b} + \vec{c}}{3}\right) \cdot \left(\frac{\vec{a} + \vec{c}}{3} - \vec{b}\right) = 0

(a⃗+b⃗+c⃗3)⋅(a⃗+c⃗−3b⃗3)=0\left(\frac{\vec{a} + \vec{b} + \vec{c}}{3}\right) \cdot \left(\frac{\vec{a} + \vec{c} - 3\vec{b}}{3}\right) = 0

19(a⃗+b⃗+c⃗)⋅(a⃗+c⃗−3b⃗)=0\frac{1}{9} (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{c} - 3\vec{b}) = 0

(a⃗+b⃗+c⃗)⋅(a⃗+c⃗−3b⃗)=0(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{c} - 3\vec{b}) = 0

Expand the dot product:

a⃗⋅(a⃗+c⃗−3b⃗)+b⃗⋅(a⃗+c⃗−3b⃗)+c⃗⋅(a⃗+c⃗−3b⃗)=0\vec{a} \cdot (\vec{a} + \vec{c} - 3\vec{b}) + \vec{b} \cdot (\vec{a} + \vec{c} - 3\vec{b}) + \vec{c} \cdot (\vec{a} + \vec{c} - 3\vec{b}) = 0

(a⃗⋅a⃗+a⃗⋅c⃗−3a⃗⋅b⃗)+(b⃗⋅a⃗+b⃗⋅c⃗−3b⃗⋅b⃗)+(c⃗⋅a⃗+c⃗⋅c⃗−3c⃗⋅b⃗)=0(\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{c} - 3\vec{a} \cdot \vec{b}) + (\vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{c} - 3\vec{b} \cdot \vec{b}) + (\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{c} - 3\vec{c} \cdot \vec{b}) = 0

∣a⃗∣2+a⃗⋅c⃗−3a⃗⋅b⃗+a⃗⋅b⃗+b⃗⋅c⃗−3∣b⃗∣2+a⃗⋅c⃗+∣c⃗∣2−3b⃗⋅c⃗=0|\vec{a}|^2 + \vec{a} \cdot \vec{c} - 3\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} - 3|\vec{b}|^2 + \vec{a} \cdot \vec{c} + |\vec{c}|^2 - 3\vec{b} \cdot \vec{c} = 0

∣a⃗∣2+∣c⃗∣2−3∣b⃗∣2+2a⃗⋅c⃗−2a⃗⋅b⃗−2b⃗⋅c⃗=0|\vec{a}|^2 + |\vec{c}|^2 - 3|\vec{b}|^2 + 2\vec{a} \cdot \vec{c} - 2\vec{a} \cdot \vec{b} - 2\vec{b} \cdot \vec{c} = 0

Substitute the values of ∣a⃗∣2|\vec{a}|^2, ∣b⃗∣2|\vec{b}|^2, ∣c⃗∣2|\vec{c}|^2 and the expressions for the dot products:

a2+c2−3b2+(a2+c2−b2)−(a2+b2−c2)−(b2+c2−a2)=0a^2 + c^2 - 3b^2 + (a^2 + c^2 - b^2) - (a^2 + b^2 - c^2) - (b^2 + c^2 - a^2) = 0

a2+c2−3b2+a2+c2−b2−a2−b2+c2−b2−c2+a2=0a^2 + c^2 - 3b^2 + a^2 + c^2 - b^2 - a^2 - b^2 + c^2 - b^2 - c^2 + a^2 = 0

Combine terms:

(a2+a2−a2+a2)+(−3b2−b2−b2−b2)+(c2+c2+c2−c2)=0(a^2 + a^2 - a^2 + a^2) + (-3b^2 - b^2 - b^2 - b^2) + (c^2 + c^2 + c^2 - c^2) = 0

2a2−6b2+2c2=02a^2 - 6b^2 + 2c^2 = 0

Divide by 2:

a2−3b2+c2=0a^2 - 3b^2 + c^2 = 0

Rearrange the terms to find the required ratio:

a2+c2=3b2a^2 + c^2 = 3b^2

a2+c2b2=3\frac{a^2 + c^2}{b^2} = 3