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

Concept: The altitude from a vertex of a triangle is the perpendicular distance from that vertex to the opposite side. For a triangle in three-dimensional geometry: \[ \text{Area of triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} \] If vectors are used, then area can also be written as: \[ \text{Area} = \frac{1}{2}|\vec{AB}\times \vec{AC}| \] Hence, \[ \text{Height from }A = \frac{|\vec{AB}\times \vec{AC}|}{|BC|} \] This formula is extremely useful in vector geometry problems involving distances and perpendiculars.

Step 1: Find the vectors \(\vec{AB}\) and \(\vec{AC}\).
Given: \[ A(1,1,1), \quad B(2,3,4), \quad C(2,5,7) \] Therefore, \[ \vec{AB} = (2-1)\hat{i}+(3-1)\hat{j}+(4-1)\hat{k} = \hat{i}+2\hat{j}+3\hat{k} \] and \[ \vec{AC} = (2-1)\hat{i}+(5-1)\hat{j}+(7-1)\hat{k} = \hat{i}+4\hat{j}+6\hat{k} \]

Step 2: Find the cross product \(\vec{AB}\times \vec{AC}\).
\[ \vec{AB}\times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 2 & 3\\ 1 & 4 & 6 \end{vmatrix} \] Expanding the determinant: \[ = \hat{i}(12-12) - \hat{j}(6-3) + \hat{k}(4-2) \] \[ = 0\hat{i}-3\hat{j}+2\hat{k} \] Hence, \[ \vec{AB}\times \vec{AC} = -3\hat{j}+2\hat{k} \] Now calculate its magnitude: \[ |\vec{AB}\times \vec{AC}| = \sqrt{0^2+(-3)^2+2^2} = \sqrt{9+4} = \sqrt{13} \]

Step 3: Find the length of side \(BC\).
\[ \vec{BC} = (2-2)\hat{i}+(5-3)\hat{j}+(7-4)\hat{k} = 0\hat{i}+2\hat{j}+3\hat{k} \] Thus, \[ |BC| = \sqrt{0^2+2^2+3^2} = \sqrt{4+9} = \sqrt{13} \]

Step 4: Compute the altitude from vertex \(A\).
Using: \[ \text{Altitude from }A = \frac{|\vec{AB}\times \vec{AC}|}{|BC|} \] Substituting the values: \[ = \frac{\sqrt{13}}{\sqrt{13}} = 1 \] Hence, the required altitude length is: \[ \boxed{1} \]