Area of the triangle whose base is given by `bar(A) =i + j + k` and one of the two sides by `bar(B) =4j + 3k` is

Options

A. `sqrt26`
B. `(sqrt26)/(2)`
C. `(5)/(2)`
D. 5

Answer

To find the area of the triangle whose base is represented by the vector \(\bar{A} = \hat{i} + \hat{j} + \hat{k}\) and one of the sides by the vector \(\bar{B} = 4\hat{j} + 3\hat{k}\), we can follow these steps:

### Step 1: Identify the vectors
We have:
- Base vector: \(\bar{A} = \hat{i} + \hat{j} + \hat{k}\)
- Side vector: \(\bar{B} = 4\hat{j} + 3\hat{k}\)

### Step 2: Calculate the cross product of the vectors
The area of the triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} |\bar{A} \times \bar{B}| \]
To find \(\bar{A} \times \bar{B}\), we set up the determinant:
\[ \bar{A} \times \bar{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 4 & 3 \end{vmatrix} \]

### Step 3: Evaluate the determinant
Calculating the determinant:
\[ \bar{A} \times \bar{B} = \hat{i} \begin{vmatrix} 1 & 1 \\ 4 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 0 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 1 & 1 \\ 4 & 3 \end{vmatrix} = (1)(3) - (1)(4) = 3 - 4 = -1\)
2. \(\begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} = (1)(3) - (1)(0) = 3 - 0 = 3\)
3. \(\begin{vmatrix} 1 & 1 \\ 0 & 4 \end{vmatrix} = (1)(4) - (1)(0) = 4 - 0 = 4\)

Putting it all together:
\[ \bar{A} \times \bar{B} = -1\hat{i} - 3\hat{j} + 4\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now, we calculate the magnitude:
\[ |\bar{A} \times \bar{B}| = \sqrt{(-1)^2 + (-3)^2 + (4)^2} = \sqrt{1 + 9 + 16} = \sqrt{26} \]

### Step 5: Calculate the area of the triangle
Finally, we find the area of the triangle:
\[ \text{Area} = \frac{1}{2} |\bar{A} \times \bar{B}| = \frac{1}{2} \sqrt{26} \]

### Final Answer
Thus, the area of the triangle is \(\frac{\sqrt{26}}{2}\).

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