Let ABC be a triangle, the position vectors of whose vertices are respectively `7 hatj + 10 hatk , -hati + 6 hat j + 6 hatk and -4 hati + 9 hatj + 6 hat k . " Then, " Delta ABC ` is

Options

A. isosceles
B. equilateral
C. right angled
D. none of these

Explanation

To solve the problem, we need to find the area of triangle ABC given the position vectors of its vertices A, B, and C. The position vectors are:

- A = \( 7\hat{j} + 10\hat{k} \)
- B = \( -\hat{i} + 6\hat{j} + 6\hat{k} \)
- C = \( -4\hat{i} + 9\hat{j} + 6\hat{k} \)

### Step 1: Find the position vectors of the vertices.

The position vectors are already given as:
- \( \vec{A} = 0\hat{i} + 7\hat{j} + 10\hat{k} \)
- \( \vec{B} = -1\hat{i} + 6\hat{j} + 6\hat{k} \)
- \( \vec{C} = -4\hat{i} + 9\hat{j} + 6\hat{k} \)

### Step 2: Calculate the vectors AB and AC.

We can find the vectors \( \vec{AB} \) and \( \vec{AC} \) using the position vectors:

\[ \vec{AB} = \vec{B} - \vec{A} = (-1\hat{i} + 6\hat{j} + 6\hat{k}) - (0\hat{i} + 7\hat{j} + 10\hat{k}) = -1\hat{i} - 1\hat{j} - 4\hat{k} \]

\[ \vec{AC} = \vec{C} - \vec{A} = (-4\hat{i} + 9\hat{j} + 6\hat{k}) - (0\hat{i} + 7\hat{j} + 10\hat{k}) = -4\hat{i} + 2\hat{j} - 4\hat{k} \]

### Step 3: Calculate the cross product \( \vec{AB} \times \vec{AC} \).

To find the area of triangle ABC, we need the magnitude of the cross product of vectors \( \vec{AB} \) and \( \vec{AC} \):

\[ \vec{AB} = -1\hat{i} - 1\hat{j} - 4\hat{k} \]
\[ \vec{AC} = -4\hat{i} + 2\hat{j} - 4\hat{k} \]

The cross product \( \vec{AB} \times \vec{AC} \) can be calculated using the determinant of a matrix:

\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & -4 \\ -4 & 2 & -4 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} -1 & -4 \\ 2 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & -4 \\ -4 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & -1 \\ -4 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ (-1)(-4) - (-4)(2) = 4 + 8 = 12 \]

2. For \( \hat{j} \):
\[ (-1)(-4) - (-4)(-4) = 4 - 16 = -12 \]

3. For \( \hat{k} \):
\[ (-1)(2) - (-1)(-4) = -2 - 4 = -6 \]

Putting it all together:

\[ \vec{AB} \times \vec{AC} = 12\hat{i} + 12\hat{j} - 6\hat{k} \]

### Step 4: Calculate the magnitude of the cross product.

Now, we find the magnitude of the cross product:

\[ |\vec{AB} \times \vec{AC}| = \sqrt{12^2 + 12^2 + (-6)^2} = \sqrt{144 + 144 + 36} = \sqrt{324} = 18 \]

### Step 5: Calculate the area of triangle ABC.

The area \( \Delta \) of triangle ABC is given by:

\[ \Delta = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \times 18 = 9 \]

Thus, the area of triangle ABC is \( \Delta = 9 \).