If a vector ofmagnitude 50 is collinear with vector `vecb = 6 hat i - 8 hat j-15/2 hat k` and makes an acute anlewih positive z-axis then:

Options

A. `24 hati - 32 hatj - 30 hatk `
B. ` -24 hati + 32 hatj + 30 hatk `
C. `12 hati - 16 hatj - 15 hatk `
D. none of these

Answer

To solve this problem step by step, we need to find a vector \( \vec{a} \) that is collinear with the vector \( \vec{b} = 6 \hat{i} - 8 \hat{j} - \frac{15}{2} \hat{k} \), has a magnitude of 50, and makes an acute angle with the positive z-axis.

### Step 1: Find the magnitude of vector \( \vec{b} \)

The magnitude of vector \( \vec{b} \) can be calculated using the formula:

\[ |\vec{b}| = \sqrt{(6)^2 + (-8)^2 + \left(-\frac{15}{2}\right)^2} \]

Calculating each term:

\[ |\vec{b}| = \sqrt{36 + 64 + \left(\frac{225}{4}\right)} \]

Converting to a common denominator:

\[ |\vec{b}| = \sqrt{36 + 64 + 56.25} = \sqrt{156.25} = 12.5 \]

### Step 2: Express vector \( \vec{a} \) in terms of \( \vec{b} \)

Since \( \vec{a} \) is collinear with \( \vec{b} \), we can express \( \vec{a} \) as:

\[ \vec{a} = \lambda \vec{b} \]

where \( \lambda \) is a scalar.

### Step 3: Set the magnitude of vector \( \vec{a} \)

Given that the magnitude of \( \vec{a} \) is 50, we have:

\[ |\vec{a}| = |\lambda| |\vec{b}| = 50 \]

Substituting the magnitude of \( \vec{b} \):

\[ |\lambda| \cdot 12.5 = 50 \]

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

Now we can solve for \( |\lambda| \):

\[ |\lambda| = \frac{50}{12.5} = 4 \]

Thus, \( \lambda \) can be either \( 4 \) or \( -4 \).

### Step 5: Determine the direction of \( \vec{a} \)

Since \( \vec{a} \) makes an acute angle with the positive z-axis, we need to ensure that the component along the z-axis is positive. The z-component of \( \vec{b} \) is \( -\frac{15}{2} \). Therefore, if \( \lambda \) is negative, \( \vec{a} \) will point in the opposite direction, making the angle obtuse.

Thus, we choose:

\[ \lambda = 4 \]

### Step 6: Write the vector \( \vec{a} \)

Now substituting \( \lambda \) back into the equation for \( \vec{a} \):

\[ \vec{a} = 4 \vec{b} = 4(6 \hat{i} - 8 \hat{j} - \frac{15}{2} \hat{k}) = 24 \hat{i} - 32 \hat{j} - 30 \hat{k} \]

### Final Result

The vector \( \vec{a} \) is:

\[ \vec{a} = 24 \hat{i} - 32 \hat{j} - 30 \hat{k} \]