The projection of vector `veca=2hati+3hatj+2hatk` along `vecb=hati+2hatj+1hatk` is

Options

A. `(2)/(3)`
B. `(1)/(3)`
C. 2
D. `sqrt(6)`

Answer

To find the projection of vector \(\vec{A} = 2\hat{i} + 3\hat{j} + 2\hat{k}\) along vector \(\vec{B} = \hat{i} + 2\hat{j} + \hat{k}\), we can use the formula for the projection of vector \(\vec{A}\) onto vector \(\vec{B}\):

\[ \text{Projection of } \vec{A} \text{ on } \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2} \vec{B} \]

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

The dot product of two vectors \(\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) is given by:

\[ \vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 + a_3b_3 \]

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

\[ \vec{A} \cdot \vec{B} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10 \]

### Step 2: Calculate the magnitude of vector \(\vec{B}\)

The magnitude of vector \(\vec{B}\) is calculated as follows:

\[ |\vec{B}| = \sqrt{b_1^2 + b_2^2 + b_3^2} \]

For \(\vec{B} = \hat{i} + 2\hat{j} + \hat{k}\):

\[ |\vec{B}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

### Step 3: Calculate the projection of \(\vec{A}\) onto \(\vec{B}\)

Using the projection formula:

\[ \text{Projection of } \vec{A} \text{ on } \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2} \vec{B} \]

First, we need to compute \(|\vec{B}|^2\):

\[ |\vec{B}|^2 = (\sqrt{6})^2 = 6 \]

Now substituting the values into the projection formula:

\[ \text{Projection of } \vec{A} \text{ on } \vec{B} = \frac{10}{6} \vec{B} = \frac{5}{3} \vec{B} \]

Substituting \(\vec{B}\):

\[ \text{Projection of } \vec{A} \text{ on } \vec{B} = \frac{5}{3} (\hat{i} + 2\hat{j} + \hat{k}) = \frac{5}{3}\hat{i} + \frac{10}{3}\hat{j} + \frac{5}{3}\hat{k} \]

### Final Answer

The projection of vector \(\vec{A}\) along vector \(\vec{B}\) is:

\[ \frac{5}{3}\hat{i} + \frac{10}{3}\hat{j} + \frac{5}{3}\hat{k} \]

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