\( A(1,2,3) \), \( B(3,4,k) \), \( C(2,1,4) \) form an isosceles triangle. If \( AB=BC \), then the area of \( \triangle ABC \) is
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For an isosceles triangle with \( AB=BC \), the midpoint of side \( AC \) forms a right angle with vertex \( B \). Calculating the base length and height using this geometric property is an elegant alternative to computing the vector cross product.
Step 1: Use the equal-side condition to find $k$. We are told $AB=BC$, so $AB^2=BC^2$. Step 2: Write each squared length. \[ AB^2=(3-1)^2+(4-2)^2+(k-3)^2=8+(k-3)^2 \] \[ BC^2=(2-3)^2+(1-4)^2+(4-k)^2=10+(4-k)^2 \] Step 3: Solve for $k$. \[ 8+(k-3)^2=10+(4-k)^2 \] Expanding: $8+k^2-6k+9=10+16-8k+k^2$, which gives $2k=9$, so $k=\tfrac{9}{2}$. Step 4: Form two side vectors from $A$. With $B=(3,4,\tfrac{9}{2})$: \[ \vec{AB}=2\hat{i}+2\hat{j}+\tfrac{3}{2}\hat{k},\qquad\vec{AC}=\hat{i}-\hat{j}+\hat{k} \] Step 5: Take the cross product. \[ \vec{AB}\times\vec{AC}=\left(2+\tfrac{3}{2}\right)\hat{i}-\left(2-\tfrac{3}{2}\right)\hat{j}+(-2-2)\hat{k}=\tfrac{7}{2}\hat{i}-\tfrac{1}{2}\hat{j}-4\hat{k} \] Step 6: Area is half the magnitude. \[ |\vec{AB}\times\vec{AC}|=\sqrt{\tfrac{49}{4}+\tfrac{1}{4}+16}=\sqrt{\tfrac{114}{4}}=\tfrac{\sqrt{114}}{2} \] \[ \text{Area}=\tfrac{1}{2}\cdot\tfrac{\sqrt{114}}{2}=\boxed{\tfrac{\sqrt{114}}{4}} \]