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Question: The position vectors of the vertices A, B and C of a tetrahedron ABCD are $\hat{i}+\hat{j}+\hat{k}, ...

The position vectors of the vertices A, B and C of a tetrahedron ABCD are i^+j^+k^,i^,and3i^\hat{i}+\hat{j}+\hat{k}, \hat{i}, and 3\hat{i} respectively. The altitude form vertex D to the opposite face ABC meets the median line through A of the triangle ABC at a point E. If the length of edge AD is 4 units and volume of tetrahedron is 223\frac{2\sqrt{2}}{3} units, then the possible position vectors (s) of point E is/are

A

−i^+3j^+3k^-\hat{i}+3\hat{j}+3\hat{k}

B

2j^+2k^2\hat{j}+2\hat{k}

C

3i^+j^+k^3\hat{i}+\hat{j}+\hat{k}

D

3i^−j^−k^3\hat{i}-\hat{j}-\hat{k}

Answer

(A), (D)

Explanation

Solution

Let the position vectors of the vertices A, B, C, and D be a⃗,b⃗,c⃗,andd⃗\vec{a}, \vec{b}, \vec{c}, and \vec{d} respectively.

Given: a⃗=i^+j^+k^\vec{a} = \hat{i}+\hat{j}+\hat{k} b⃗=i^\vec{b} = \hat{i} c⃗=3i^\vec{c} = 3\hat{i} Let d⃗=xi^+yj^+zk^\vec{d} = x\hat{i} + y\hat{j} + z\hat{k}.

The length of edge AD is 4 units: ∣d⃗−a⃗∣=4  ⟹  ∣(x−1)i^+(y−1)j^+(z−1)k^∣=4|\vec{d} - \vec{a}| = 4 \implies |(x-1)\hat{i} + (y-1)\hat{j} + (z-1)\hat{k}| = 4 (x−1)2+(y−1)2+(z−1)2=16(x-1)^2 + (y-1)^2 + (z-1)^2 = 16 (Equation 1)

The volume of the tetrahedron ABCD is 223\frac{2\sqrt{2}}{3}. The volume is given by V=16∣(b⃗−a⃗)×(c⃗−a⃗)⋅(d⃗−a⃗)∣V = \frac{1}{6} |(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a}) \cdot (\vec{d}-\vec{a})|. AB⃗=b⃗−a⃗=i^−(i^+j^+k^)=−j^−k^\vec{AB} = \vec{b} - \vec{a} = \hat{i} - (\hat{i}+\hat{j}+\hat{k}) = -\hat{j}-\hat{k} AC⃗=c⃗−a⃗=3i^−(i^+j^+k^)=2i^−j^−k^\vec{AC} = \vec{c} - \vec{a} = 3\hat{i} - (\hat{i}+\hat{j}+\hat{k}) = 2\hat{i}-\hat{j}-\hat{k} AB⃗×AC⃗=(−j^−k^)×(2i^−j^−k^)=−2(j^×i^)−(j^×−j^)−(j^×−k^)−(k^×2i^)−(k^×−j^)−(k^×−k^)\vec{AB} \times \vec{AC} = (-\hat{j}-\hat{k}) \times (2\hat{i}-\hat{j}-\hat{k}) = -2(\hat{j} \times \hat{i}) - (\hat{j} \times -\hat{j}) - (\hat{j} \times -\hat{k}) - (\hat{k} \times 2\hat{i}) - (\hat{k} \times -\hat{j}) - (\hat{k} \times -\hat{k}) =−2(−k^)−0⃗−(−i^)−2j^−i^−0⃗=2k^+i^−2j^−i^=−2j^+2k^= -2(-\hat{k}) - \vec{0} - (-\hat{i}) - 2\hat{j} - \hat{i} - \vec{0} = 2\hat{k} + \hat{i} - 2\hat{j} - \hat{i} = -2\hat{j} + 2\hat{k}. AD⃗=d⃗−a⃗=(x−1)i^+(y−1)j^+(z−1)k^\vec{AD} = \vec{d} - \vec{a} = (x-1)\hat{i} + (y-1)\hat{j} + (z-1)\hat{k}. (AB⃗×AC⃗)⋅AD⃗=(−2j^+2k^)⋅((x−1)i^+(y−1)j^+(z−1)k^)=−2(y−1)+2(z−1)=2z−2y(\vec{AB} \times \vec{AC}) \cdot \vec{AD} = (-2\hat{j} + 2\hat{k}) \cdot ((x-1)\hat{i} + (y-1)\hat{j} + (z-1)\hat{k}) = -2(y-1) + 2(z-1) = 2z - 2y. V=16∣2z−2y∣=13∣z−y∣V = \frac{1}{6} |2z - 2y| = \frac{1}{3} |z - y|. Given V=223V = \frac{2\sqrt{2}}{3}, so 13∣z−y∣=223  ⟹  ∣z−y∣=22\frac{1}{3} |z - y| = \frac{2\sqrt{2}}{3} \implies |z - y| = 2\sqrt{2}. Thus, z−y=22z - y = 2\sqrt{2} or z−y=−22z - y = -2\sqrt{2}. (Equation 2)

The altitude from D to face ABC is a line through D perpendicular to the plane ABC. The direction vector of this altitude is parallel to the normal vector of the plane ABC, which is N⃗=AB⃗×AC⃗=−2j^+2k^\vec{N} = \vec{AB} \times \vec{AC} = -2\hat{j} + 2\hat{k}. The equation of the plane ABC is (r⃗−a⃗)⋅N⃗=0(\vec{r} - \vec{a}) \cdot \vec{N} = 0. (r⃗−(i^+j^+k^))⋅(−2j^+2k^)=0(\vec{r} - (\hat{i}+\hat{j}+\hat{k})) \cdot (-2\hat{j} + 2\hat{k}) = 0. Let r⃗=x′i^+y′j^+z′k^\vec{r} = x'\hat{i} + y'\hat{j} + z'\hat{k}. ((x′−1)i^+(y′−1)j^+(z′−1)k^)⋅(−2j^+2k^)=0  ⟹  −2(y′−1)+2(z′−1)=0  ⟹  y′−1=z′−1  ⟹  y′=z′((x'-1)\hat{i} + (y'-1)\hat{j} + (z'-1)\hat{k}) \cdot (-2\hat{j} + 2\hat{k}) = 0 \implies -2(y'-1) + 2(z'-1) = 0 \implies y'-1 = z'-1 \implies y' = z'. The equation of the plane ABC is y=zy=z.

The point E lies on the altitude from D. So, e⃗=d⃗+kN⃗=(xi^+yj^+zk^)+k(−2j^+2k^)=xi^+(y−2k)j^+(z+2k)k^\vec{e} = \vec{d} + k\vec{N} = (x\hat{i} + y\hat{j} + z\hat{k}) + k(-2\hat{j} + 2\hat{k}) = x\hat{i} + (y-2k)\hat{j} + (z+2k)\hat{k} for some scalar kk.

The point E also lies on the median line through A of triangle ABC. Let M be the midpoint of BC. m⃗=b⃗+c⃗2=i^+3i^2=2i^\vec{m} = \frac{\vec{b}+\vec{c}}{2} = \frac{\hat{i} + 3\hat{i}}{2} = 2\hat{i}. The median line through A is the line AM. The direction vector of AM is AM⃗=m⃗−a⃗=2i^−(i^+j^+k^)=i^−j^−k^\vec{AM} = \vec{m} - \vec{a} = 2\hat{i} - (\hat{i}+\hat{j}+\hat{k}) = \hat{i}-\hat{j}-\hat{k}. The position vector of a point on the line AM is r⃗AM=a⃗+λAM⃗=(i^+j^+k^)+λ(i^−j^−k^)=(1+λ)i^+(1−λ)j^+(1−λ)k^\vec{r}_{AM} = \vec{a} + \lambda \vec{AM} = (\hat{i}+\hat{j}+\hat{k}) + \lambda (\hat{i}-\hat{j}-\hat{k}) = (1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1-\lambda)\hat{k} for some scalar λ\lambda.

Since E lies on both lines, its position vector e⃗\vec{e} must satisfy both forms: xi^+(y−2k)j^+(z+2k)k^=(1+λ)i^+(1−λ)j^+(1−λ)k^x\hat{i} + (y-2k)\hat{j} + (z+2k)\hat{k} = (1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1-\lambda)\hat{k}. Equating coefficients: x=1+λx = 1+\lambda (3) y−2k=1−λy-2k = 1-\lambda (4) z+2k=1−λz+2k = 1-\lambda (5)

From (4) and (5): y−2k=z+2k  ⟹  y−z=4k  ⟹  k=y−z4y-2k = z+2k \implies y-z = 4k \implies k = \frac{y-z}{4}. From (4) and (5): 1−λ=y−2k=y−2(y−z4)=y−y−z2=2y−y+z2=y+z21-\lambda = y-2k = y - 2\left(\frac{y-z}{4}\right) = y - \frac{y-z}{2} = \frac{2y-y+z}{2} = \frac{y+z}{2}. So, 1−λ=y+z21-\lambda = \frac{y+z}{2}. (6) From (3), λ=x−1\lambda = x-1. Substitute into (6): 1−(x−1)=y+z2  ⟹  2−x=y+z2  ⟹  4−2x=y+z1-(x-1) = \frac{y+z}{2} \implies 2-x = \frac{y+z}{2} \implies 4-2x = y+z. (7)

We have a system of equations for x, y, z: (1) (x−1)2+(y−1)2+(z−1)2=16(x-1)^2 + (y-1)^2 + (z-1)^2 = 16 (2) ∣z−y∣=22|z-y| = 2\sqrt{2} (7) y+z=4−2xy+z = 4-2x

Case 1: z−y=22z-y = 2\sqrt{2}. Adding (2) and (7): 2z=4−2x+22  ⟹  z=2−x+22z = 4-2x+2\sqrt{2} \implies z = 2-x+\sqrt{2}. Subtracting (2) from (7): 2y=4−2x−22  ⟹  y=2−x−22y = 4-2x-2\sqrt{2} \implies y = 2-x-\sqrt{2}. Substitute y and z into (1): (x−1)2+((2−x−2)−1)2+((2−x+2)−1)2=16(x-1)^2 + ((2-x-\sqrt{2})-1)^2 + ((2-x+\sqrt{2})-1)^2 = 16 (x−1)2+(1−x−2)2+(1−x+2)2=16(x-1)^2 + (1-x-\sqrt{2})^2 + (1-x+\sqrt{2})^2 = 16 Let u=1−xu = 1-x. (−(1−x))2+(u−2)2+(u+2)2=16(-(1-x))^2 + (u-\sqrt{2})^2 + (u+\sqrt{2})^2 = 16 u2+(u2−22u+2)+(u2+22u+2)=16u^2 + (u^2 - 2\sqrt{2}u + 2) + (u^2 + 2\sqrt{2}u + 2) = 16 3u2+4=16  ⟹  3u2=12  ⟹  u2=4  ⟹  u=±23u^2 + 4 = 16 \implies 3u^2 = 12 \implies u^2 = 4 \implies u = \pm 2. If u=2u=2, 1−x=2  ⟹  x=−11-x=2 \implies x=-1. Then y=2−(−1)−2=3−2y = 2-(-1)-\sqrt{2} = 3-\sqrt{2}, z=2−(−1)+2=3+2z = 2-(-1)+\sqrt{2} = 3+\sqrt{2}. If u=−2u=-2, 1−x=−2  ⟹  x=31-x=-2 \implies x=3. Then y=2−3−2=−1−2y = 2-3-\sqrt{2} = -1-\sqrt{2}, z=2−3+2=−1+2z = 2-3+\sqrt{2} = -1+\sqrt{2}.

Case 2: z−y=−22z-y = -2\sqrt{2}. Adding (2) and (7): 2z=4−2x−22  ⟹  z=2−x−22z = 4-2x-2\sqrt{2} \implies z = 2-x-\sqrt{2}. Subtracting (2) from (7): 2y=4−2x+22  ⟹  y=2−x+22y = 4-2x+2\sqrt{2} \implies y = 2-x+\sqrt{2}. Substitute y and z into (1): (x−1)2+((2−x+2)−1)2+((2−x−2)−1)2=16(x-1)^2 + ((2-x+\sqrt{2})-1)^2 + ((2-x-\sqrt{2})-1)^2 = 16 (x−1)2+(1−x+2)2+(1−x−2)2=16(x-1)^2 + (1-x+\sqrt{2})^2 + (1-x-\sqrt{2})^2 = 16 This is the same equation as in Case 1, 3u2+4=163u^2 + 4 = 16, which gives u=±2u = \pm 2. If u=2u=2, 1−x=2  ⟹  x=−11-x=2 \implies x=-1. Then y=2−(−1)+2=3+2y = 2-(-1)+\sqrt{2} = 3+\sqrt{2}, z=2−(−1)−2=3−2z = 2-(-1)-\sqrt{2} = 3-\sqrt{2}. If u=−2u=-2, 1−x=−2  ⟹  x=31-x=-2 \implies x=3. Then y=2−3+2=−1+2y = 2-3+\sqrt{2} = -1+\sqrt{2}, z=2−3−2=−1−2z = 2-3-\sqrt{2} = -1-\sqrt{2}.

So, the possible position vectors for D are: D1:x=−1,y=3−2,z=3+2D_1: x=-1, y=3-\sqrt{2}, z=3+\sqrt{2} D2:x=3,y=−1−2,z=−1+2D_2: x=3, y=-1-\sqrt{2}, z=-1+\sqrt{2} D3:x=−1,y=3+2,z=3−2D_3: x=-1, y=3+\sqrt{2}, z=3-\sqrt{2} D4:x=3,y=−1+2,z=−1−2D_4: x=3, y=-1+\sqrt{2}, z=-1-\sqrt{2}

Now we find the position vector of E, e⃗=(1+λ)i^+(1−λ)j^+(1−λ)k^\vec{e} = (1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1-\lambda)\hat{k}, where λ=x−1\lambda = x-1. For D1D_1: x=−1  ⟹  λ=−1−1=−2x=-1 \implies \lambda = -1-1 = -2. e⃗1=(1−2)i^+(1−(−2))j^+(1−(−2))k^=−i^+3j^+3k^\vec{e}_1 = (1-2)\hat{i} + (1-(-2))\hat{j} + (1-(-2))\hat{k} = -\hat{i} + 3\hat{j} + 3\hat{k}. For D2D_2: x=3  ⟹  λ=3−1=2x=3 \implies \lambda = 3-1 = 2. e⃗2=(1+2)i^+(1−2)j^+(1−2)k^=3i^−j^−k^\vec{e}_2 = (1+2)\hat{i} + (1-2)\hat{j} + (1-2)\hat{k} = 3\hat{i} - \hat{j} - \hat{k}. For D3D_3: x=−1  ⟹  λ=−1−1=−2x=-1 \implies \lambda = -1-1 = -2. e⃗3=(1−2)i^+(1−(−2))j^+(1−(−2))k^=−i^+3j^+3k^\vec{e}_3 = (1-2)\hat{i} + (1-(-2))\hat{j} + (1-(-2))\hat{k} = -\hat{i} + 3\hat{j} + 3\hat{k}. For D4D_4: x=3  ⟹  λ=3−1=2x=3 \implies \lambda = 3-1 = 2. e⃗4=(1+2)i^+(1−2)j^+(1−2)k^=3i^−j^−k^\vec{e}_4 = (1+2)\hat{i} + (1-2)\hat{j} + (1-2)\hat{k} = 3\hat{i} - \hat{j} - \hat{k}.

The possible position vectors of point E are −i^+3j^+3k^-\hat{i} + 3\hat{j} + 3\hat{k} and 3i^−j^−k^3\hat{i} - \hat{j} - \hat{k}.

Comparing these with the given options: (A) −i^+3j^+3k^-\hat{i}+3\hat{j}+3\hat{k} - Matches one of the possible position vectors. (B) 2j^+2k^2\hat{j}+2\hat{k} - Does not match. (C) 3i^+j^+k^3\hat{i}+\hat{j}+\hat{k} - Does not match. (D) 3i^−j^−k^3\hat{i}-\hat{j}-\hat{k} - Matches one of the possible position vectors.

Both options (A) and (D) are possible position vectors for point E.