Question

If \(A(1,1,1)\), \(B(2,3,4)\) and \(C(2,5,7)\) are the vertices of \(\triangle ABC\), then the length of the altitude drawn through the vertex \(A\) is:

• A. \(2\)
• B. \(1\)
• C. \(\sqrt{2}\)
• D. \(\sqrt{3}\)

Hint:

For triangles in 3D geometry: \[ \text{Altitude from a vertex} = \frac{|\vec{AB}\times \vec{AC}|}{|BC|} \] Always remember: \[ \text{Area} = \frac12 |\vec{AB}\times \vec{AC}| \] which directly connects vector algebra with geometric distances.

Answer 1

The Correct Option is B

Step 1: Plan with the area method.
The altitude from $A$ is the height of the triangle when $BC$ is the base. Using vectors, \[ \text{height from }A=\frac{|\vec{AB}\times\vec{AC}|}{|BC|}. \]
Step 2: Find the two side vectors from $A$.
With $A(1,1,1)$, $B(2,3,4)$, $C(2,5,7)$: \[ \vec{AB}=\hat i+2\hat j+3\hat k,\qquad \vec{AC}=\hat i+4\hat j+6\hat k. \]
Step 3: Take the cross product.
\[ \vec{AB}\times\vec{AC}=\hat i(2\cdot6-3\cdot4)-\hat j(1\cdot6-3\cdot1)+\hat k(1\cdot4-2\cdot1)=0\hat i-3\hat j+2\hat k. \]
Step 4: Find its length (twice the area).
\[ |\vec{AB}\times\vec{AC}|=\sqrt{0+9+4}=\sqrt{13}. \]
Step 5: Find the base length $BC$.
$\vec{BC}=2\hat j+3\hat k$, so $|BC|=\sqrt{0+4+9}=\sqrt{13}$.
Step 6: Divide to get the altitude.
\[ \text{altitude}=\frac{\sqrt{13}}{\sqrt{13}}=1. \] \[ \boxed{1} \]