The vector `3i-2j+k,i-3j+5k and 2i+j-4k` form the sides of a triangle, This triangle is

Options

A. an acute angled triangle
B. an obtuse angled triangle
C. a right angled triangle
D. an equilateral triangle

Answer

To determine the type of triangle formed by the vectors \( \vec{A} = 3\hat{i} - 2\hat{j} + \hat{k} \), \( \vec{B} = \hat{i} - 3\hat{j} + 5\hat{k} \), and \( \vec{C} = 2\hat{i} + \hat{j} - 4\hat{k} \), we will find the magnitudes of these vectors and check if they satisfy the Pythagorean theorem.

### Step 1: Calculate the magnitude of vector \( \vec{A} \)

The magnitude of a vector \( \vec{A} = a\hat{i} + b\hat{j} + c\hat{k} \) is given by:

\[ |\vec{A}| = \sqrt{a^2 + b^2 + c^2} \]

For \( \vec{A} = 3\hat{i} - 2\hat{j} + \hat{k} \):

\[ |\vec{A}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \]

### Step 2: Calculate the magnitude of vector \( \vec{B} \)

For \( \vec{B} = \hat{i} - 3\hat{j} + 5\hat{k} \):

\[ |\vec{B}| = \sqrt{1^2 + (-3)^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35} \]

### Step 3: Calculate the magnitude of vector \( \vec{C} \)

For \( \vec{C} = 2\hat{i} + \hat{j} - 4\hat{k} \):

\[ |\vec{C}| = \sqrt{2^2 + 1^2 + (-4)^2} = \sqrt{4 + 1 + 16} = \sqrt{21} \]

### Step 4: Check if the triangle is a right triangle using the Pythagorean theorem

The Pythagorean theorem states that for a right triangle with sides \( a \), \( b \), and \( c \) (where \( c \) is the longest side):

\[ c^2 = a^2 + b^2 \]

Let’s assign the magnitudes we calculated:

- \( |\vec{A}|^2 = 14 \)
- \( |\vec{B}|^2 = 35 \)
- \( |\vec{C}|^2 = 21 \)

Now, we check if \( |\vec{B}|^2 = |\vec{A}|^2 + |\vec{C}|^2 \):

\[ 35 = 14 + 21 \]

Since this equality holds true, we can conclude that the triangle formed by the vectors is a right triangle.

### Conclusion

The triangle formed by the vectors \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) is a right triangle.

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