Dot product of two vectors `overset(rarr)A` and `overset(rarr)B` is defined as `overset(rarr)A.overset(rarr)B=aB cos phi` , where `phi` is angle between them when they are drawn with tails coinciding. For any two vectors . This means `overset(rarr)A . overset(rarr)B=overset(rarr)B. overset(rarr)A` that . The scalar product obeys the commutative law of multiplication, the order of the two vectors does not matter. The vector product of two vectors `overset(rarr)A` and `overset(rarr)B` also called the cross product, is denoted by `overset(rarr)A xx overset(rarr)B` . As the name suggests, the vector product is itself a vector. `overset(rarr)C=overset(rarr)A xx overset(rarr)B` then `C=AB sin theta` ,
`overset(rarr)A=hat i+ hat j-hatk` and `overset(rarr)B=2 hat i +3 hat j +5 hat k` angle between `overset(rarr)A` and `overset(rarr)B` is

Options

A. `120^(@)`
B. `90^(@)`
C. `60^(@)`
D. `30^(@)`

Explanation

To find the angle between the two vectors \(\overset{\rarr}{A}\) and \(\overset{\rarr}{B}\), we can use the dot product formula. Here are the steps to solve the problem:

### Step 1: Write down the vectors
Given:
\[ \overset{\rarr}{A} = \hat{i} + \hat{j} - \hat{k} \]
\[ \overset{\rarr}{B} = 2\hat{i} + 3\hat{j} + 5\hat{k} \]

### Step 2: Calculate the dot product \(\overset{\rarr}{A} \cdot \overset{\rarr}{B}\)
The dot product is calculated as follows:
\[ \overset{\rarr}{A} \cdot \overset{\rarr}{B} = (1)(2) + (1)(3) + (-1)(5) \]
Calculating this gives:
\[ = 2 + 3 - 5 = 0 \]

### Step 3: Calculate the magnitudes of the vectors
The magnitude of \(\overset{\rarr}{A}\) is:
\[ |\overset{\rarr}{A}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

The magnitude of \(\overset{\rarr}{B}\) is:
\[ |\overset{\rarr}{B}| = \sqrt{2^2 + 3^2 + 5^2} = \sqrt{4 + 9 + 25} = \sqrt{38} \]

### Step 4: Use the dot product to find the angle
The dot product formula is given by:
\[ \overset{\rarr}{A} \cdot \overset{\rarr}{B} = |\overset{\rarr}{A}| |\overset{\rarr}{B}| \cos \phi \]
Substituting the values we have:
\[ 0 = \sqrt{3} \cdot \sqrt{38} \cdot \cos \phi \]

### Step 5: Solve for \(\cos \phi\)
Since the left-hand side is zero, we can conclude:
\[ \cos \phi = 0 \]

### Step 6: Determine the angle \(\phi\)
The angle \(\phi\) for which \(\cos \phi = 0\) is:
\[ \phi = 90^\circ \]

### Final Answer
Thus, the angle between the vectors \(\overset{\rarr}{A}\) and \(\overset{\rarr}{B}\) is:
\[ \phi = 90^\circ \]

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