The angle between the two vectors, `(vec(A)=3hat(i)+4hatj+5hat(k)) and (vec(B)=3hat(i)+4hat(j)-5hat(k))` will be

Options

A. zero
B. `45^(@)`
C. `90^(@)`
D. `180^(@)`

Answer

To find the angle between the two vectors \(\vec{A} = 3\hat{i} + 4\hat{j} + 5\hat{k}\) and \(\vec{B} = 3\hat{i} + 4\hat{j} - 5\hat{k}\), we can use the dot product formula. The angle \(\theta\) between two vectors can be calculated using the formula:

\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]

### Step 1: Calculate the dot product \(\vec{A} \cdot \vec{B}\)

The dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is calculated as follows:

\[ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z \]

Substituting the components of \(\vec{A}\) and \(\vec{B}\):

\[ \vec{A} \cdot \vec{B} = (3)(3) + (4)(4) + (5)(-5) \]

Calculating each term:

\[ = 9 + 16 - 25 = 0 \]

### Step 2: Calculate the magnitudes of \(\vec{A}\) and \(\vec{B}\)

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

\[ |\vec{A}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \]

Calculating the magnitude of \(\vec{A}\):

\[ |\vec{A}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \]

Calculating the magnitude of \(\vec{B}\):

\[ |\vec{B}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \]

### Step 3: Use the dot product to find \(\cos \theta\)

Now, substituting the values into the dot product formula:

\[ 0 = (5\sqrt{2})(5\sqrt{2}) \cos \theta \]

This simplifies to:

\[ 0 = 50 \cos \theta \]

### Step 4: Solve for \(\theta\)

Since \(50 \cos \theta = 0\), we have:

\[ \cos \theta = 0 \]

The angle \(\theta\) for which \(\cos \theta = 0\) is:

\[ \theta = 90^\circ \]

### Final Answer

The angle between the two vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).

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