A particle moves `21m` along the vector `6hat(i)+2hat(j)+3hat(k)` , then `14 m` along the vector `3hat(i)-2hat(j)+6hat(k)`. Its total displacement (in meters) is

Options

A. `15hat(i)+2hat(j)+12hat(k)`
B. `9hat(i)+12hat(k)`
C. `9hat(i)+6hat(j)`
D. `24hat(i)+2hat(j)+21hat(k)`

Explanation

To find the total displacement of the particle, we need to break down the movements along the given vectors and then combine them.

### Step 1: Calculate the unit vector for the first displacement
The first displacement is along the vector \( \vec{A} = 6\hat{i} + 2\hat{j} + 3\hat{k} \).
First, we need to find the magnitude of this vector:

\[ |\vec{A}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \]

Now, the unit vector \( \hat{A} \) in the direction of \( \vec{A} \) is given by:

\[ \hat{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{7} = \frac{6}{7}\hat{i} + \frac{2}{7}\hat{j} + \frac{3}{7}\hat{k} \]

### Step 2: Calculate the displacement vector for the first movement
The particle moves \( 21 \, \text{m} \) in the direction of \( \hat{A} \):

\[ \vec{D_1} = 21 \times \hat{A} = 21 \left( \frac{6}{7}\hat{i} + \frac{2}{7}\hat{j} + \frac{3}{7}\hat{k} \right) = 18\hat{i} + 6\hat{j} + 9\hat{k} \]

### Step 3: Calculate the unit vector for the second displacement
The second displacement is along the vector \( \vec{B} = 3\hat{i} - 2\hat{j} + 6\hat{k} \).
First, we find the magnitude of this vector:

\[ |\vec{B}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \]

Now, the unit vector \( \hat{B} \) in the direction of \( \vec{B} \) is given by:

\[ \hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7} = \frac{3}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k} \]

### Step 4: Calculate the displacement vector for the second movement
The particle moves \( 14 \, \text{m} \) in the direction of \( \hat{B} \):

\[ \vec{D_2} = 14 \times \hat{B} = 14 \left( \frac{3}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k} \right) = 6\hat{i} - 4\hat{j} + 12\hat{k} \]

### Step 5: Calculate the total displacement vector
Now, we can find the total displacement \( \vec{D} \) by adding \( \vec{D_1} \) and \( \vec{D_2} \):

\[ \vec{D} = \vec{D_1} + \vec{D_2} = (18\hat{i} + 6\hat{j} + 9\hat{k}) + (6\hat{i} - 4\hat{j} + 12\hat{k}) \]

Combining the components:

\[ \vec{D} = (18 + 6)\hat{i} + (6 - 4)\hat{j} + (9 + 12)\hat{k} = 24\hat{i} + 2\hat{j} + 21\hat{k} \]

### Step 6: Write the final answer
Thus, the total displacement of the particle is:

\[ \vec{D} = 24\hat{i} + 2\hat{j} + 21\hat{k} \, \text{meters} \]

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