Question

The altitude through vertex A of $\triangle ABC$ with position vectors of points A, B, C as $\vec{a}, \vec{b}, \vec{c}$ respectively is ______.

• A. $|\vec{b} \times \vec{c}| / |\vec{c} - \vec{b}|$
• B. $|\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| / |\vec{c} - \vec{b}|$
• C. $|\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| / |\vec{c} \times \vec{b}|$
• D. $|\vec{b} \times \vec{c}| / |\vec{a}|$

Hint:

The vector quantity $(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})$ is incredibly important. Its magnitude directly represents exactly twice the area of the triangle formed by the three position vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the length of the altitude drawn from vertex $A$ to the opposite side $BC$ using vector algebra.

Step 2: Key Formula or Approach:
Geometrically, the altitude $h$ is the perpendicular distance from point $A$ to the line passing through points $B$ and $C$.
Alternatively, we can use the area of the triangle.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} |\vec{BC}| \times h$.
We also know the area in terms of position vectors is $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.

Step 3: Detailed Explanation:
Let's express the vectors:
The base vector is $\vec{BC} = \vec{c} - \vec{b}$. Its length is $|\vec{c} - \vec{b}|$.
The vector $\vec{AB} = \vec{b} - \vec{a}$ and vector $\vec{AC} = \vec{c} - \vec{a}$.
The area of the triangle is given by the cross product:
$$\text{Area} = \frac{1}{2} |(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})|$$
Expand the cross product algebraically:
$$(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a}$$
Since $\vec{a} \times \vec{a} = 0$, and reversing the order of a cross product flips the sign ($-\vec{b} \times \vec{a} = \vec{a} \times \vec{b}$ and $-\vec{a} \times \vec{c} = \vec{c} \times \vec{a}$), we get:
$$\text{Area} = \frac{1}{2} |\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a}|$$
$$\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}|$$
Now, equate this area to the standard geometric formula:
$$\frac{1}{2} |\vec{c} - \vec{b}| \times h = \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}|$$
Isolating the altitude $h$:
$$h = \frac{|\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}|}{|\vec{c} - \vec{b}|}$$

Step 4: Final Answer:
The altitude expression perfectly matches option (b).