If position vectors `vec(a)(3hati+4hatj).vec(b)(5hatj),vec( c )(4hati-3hatj)` form a triangle ABC, then position vector its orthocenter is

Options

A. `5hati`
B. `5hatk`
C. `7hati+6hatj`
D. `7hati+9hatj`

Answer

To find the position vector of the orthocenter of triangle ABC formed by the position vectors \( \vec{a} = 3\hat{i} + 4\hat{j} \), \( \vec{b} = 5\hat{j} \), and \( \vec{c} = 4\hat{i} - 3\hat{j} \), we can follow these steps:

### Step 1: Identify the coordinates of the points A, B, and C
- From the position vectors:
- Point A: \( (3, 4) \)
- Point B: \( (0, 5) \)
- Point C: \( (4, -3) \)

### Step 2: Calculate the slopes of the sides of the triangle
- Slope of line AB:
\[ \text{slope}_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 4}{0 - 3} = \frac{1}{-3} = -\frac{1}{3} \]

- Slope of line BC:
\[ \text{slope}_{BC} = \frac{-3 - 5}{4 - 0} = \frac{-8}{4} = -2 \]

- Slope of line CA:
\[ \text{slope}_{CA} = \frac{4 - (-3)}{3 - 4} = \frac{7}{-1} = -7 \]

### Step 3: Find the slopes of the altitudes
- The slope of the altitude from point A (perpendicular to BC):
\[ \text{slope}_{AD} = \frac{1}{\text{slope}_{BC}} = \frac{1}{-2} = \frac{1}{2} \]

- The slope of the altitude from point B (perpendicular to CA):
\[ \text{slope}_{BE} = \frac{1}{\text{slope}_{CA}} = \frac{1}{-7} = \frac{1}{-7} \]

### Step 4: Write the equations of the altitudes
- Equation of altitude AD (from A to BC):
Using point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - 4 = \frac{1}{2}(x - 3) \implies 2y - 8 = x - 3 \implies x - 2y + 5 = 0 \quad \text{(Equation 1)} \]

- Equation of altitude BE (from B to CA):
\[ y - 5 = -7(x - 0) \implies y - 5 = -7x \implies 7x + y - 5 = 0 \quad \text{(Equation 2)} \]

### Step 5: Solve the equations simultaneously to find the orthocenter
- Solve Equation 1 and Equation 2:
\[ x - 2y + 5 = 0 \quad (1) \]
\[ 7x + y - 5 = 0 \quad (2) \]

From Equation (1):
\[ x = 2y - 5 \]

Substituting into Equation (2):
\[ 7(2y - 5) + y - 5 = 0 \]
\[ 14y - 35 + y - 5 = 0 \]
\[ 15y - 40 = 0 \implies y = \frac{40}{15} = \frac{8}{3} \]

Substituting \( y \) back to find \( x \):
\[ x = 2\left(\frac{8}{3}\right) - 5 = \frac{16}{3} - \frac{15}{3} = \frac{1}{3} \]

### Step 6: Write the position vector of the orthocenter
The orthocenter coordinates are \( \left(\frac{1}{3}, \frac{8}{3}\right) \).
Thus, the position vector of the orthocenter is:
\[ \vec{H} = \frac{1}{3}\hat{i} + \frac{8}{3}\hat{j} \]

### Final Answer
The position vector of the orthocenter is \( \frac{1}{3}\hat{i} + \frac{8}{3}\hat{j} \).

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