If the vectors `vecA=2hati+4hatj and vecB=5hati-phatj` are parallel to each other, the magnitude of `vecB` is

Options

A. `5sqrt5`
B. `10`
C. 15
D. `2sqrt5`

Answer

To determine the magnitude of the vector \(\vec{B}\) given that the vectors \(\vec{A} = 2\hat{i} + 4\hat{j}\) and \(\vec{B} = 5\hat{i} - p\hat{j}\) are parallel, we can follow these steps:

### Step 1: Understand the condition for parallel vectors
Two vectors are parallel if their cross product is zero, i.e., \(\vec{A} \times \vec{B} = 0\).

### Step 2: Write down the vectors
We have:
\[ \vec{A} = 2\hat{i} + 4\hat{j} \]
\[ \vec{B} = 5\hat{i} - p\hat{j} \]

### Step 3: Compute the cross product \(\vec{A} \times \vec{B}\)
Using the determinant formula for the cross product:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 0 \\ 5 & -p & 0 \end{vmatrix} \]

### Step 4: Calculate the determinant
Calculating the determinant, we have:
\[ \vec{A} \times \vec{B} = \hat{i}(4 \cdot 0 - 0 \cdot (-p)) - \hat{j}(2 \cdot 0 - 0 \cdot 5) + \hat{k}(2 \cdot (-p) - 4 \cdot 5) \]
This simplifies to:
\[ \vec{A} \times \vec{B} = \hat{k}(-2p - 20) \]

### Step 5: Set the cross product to zero
For the vectors to be parallel:
\[ -2p - 20 = 0 \]
Solving for \(p\):
\[ -2p = 20 \implies p = -10 \]

### Step 6: Substitute \(p\) back into \(\vec{B}\)
Now substituting \(p = -10\) into \(\vec{B}\):
\[ \vec{B} = 5\hat{i} - (-10)\hat{j} = 5\hat{i} + 10\hat{j} \]

### Step 7: Calculate the magnitude of \(\vec{B}\)
The magnitude of \(\vec{B}\) is given by:
\[ |\vec{B}| = \sqrt{(5)^2 + (10)^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5} \]

### Final Answer
Thus, the magnitude of \(\vec{B}\) is:
\[ |\vec{B}| = 5\sqrt{5} \]

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