If `vecb ne 0`, then every vector `veca` can be written in a unique manner as the sum of a vector `veca_(p)` parallel to `vecb` and a vector `veca_(q)` perpendicular to `vecb`. If `veca` is parallel to `vecb` then `veca_(q)`=0 and `veca_(q)=veca`. The vector `veca_(p)` is called the projection of `veca` on `vecb` and is denoted by proj `vecb(veca)`. Since proj`vecb(veca)` is parallel to `vecb`, it is a scalar multiple of the vector in the direction of `vecb` i.e.,
proj `vecb(veca)=lambdavecUvecb" " (vecUvecb=(vecb)/(|vecb|))`
The scalar `lambda` is called the componennt of `veca` in the direction of `vecb` and is denoted by comp `vecb(veca)`. In fact proj `vecb(veca)=(veca.vecUvecb)vecUvecb` and comp `vecb(veca)=veca.vecUvecb`.
If `veca=-2hatj+hatj+hatk` and `vecb=4hati-3hatj+hatk` then proj `vecb(veca)` is

Options

A. (a)`4hati+3hatj+2hatk`
B. (b)`-(5)/(13)(4hati-hatj+2hatk)`
C. (c)`(5)/(13)(4hai-3hatj+hatk)`
D. (d)`-(4)/(11)(4hati-3hatj+2hatk)`

Answer

To find the projection of vector \(\vec{a}\) onto vector \(\vec{b}\), we can follow these steps:

### Step 1: Identify the vectors
Given:
\[ \vec{a} = -2\hat{j} + \hat{j} + \hat{k} = -\hat{j} + \hat{k} \]
\[ \vec{b} = 4\hat{i} - 3\hat{j} + \hat{k} \]

### Step 2: Calculate the unit vector of \(\vec{b}\)
The unit vector \(\hat{u}_{\vec{b}}\) in the direction of \(\vec{b}\) is given by:
\[ \hat{u}_{\vec{b}} = \frac{\vec{b}}{|\vec{b}|} \]
First, we need to find the magnitude of \(\vec{b}\):
\[ |\vec{b}| = \sqrt{(4)^2 + (-3)^2 + (1)^2} = \sqrt{16 + 9 + 1} = \sqrt{26} \]
Now, we can find \(\hat{u}_{\vec{b}}\):
\[ \hat{u}_{\vec{b}} = \frac{4\hat{i} - 3\hat{j} + \hat{k}}{\sqrt{26}} \]

### Step 3: Calculate the dot product \(\vec{a} \cdot \hat{u}_{\vec{b}}\)
Now we compute the dot product \(\vec{a} \cdot \hat{u}_{\vec{b}}\):
\[ \vec{a} \cdot \hat{u}_{\vec{b}} = (-\hat{j} + \hat{k}) \cdot \left(\frac{4\hat{i} - 3\hat{j} + \hat{k}}{\sqrt{26}}\right) \]
Calculating the dot product:
\[ = 0 \cdot \frac{4}{\sqrt{26}} + (-1) \cdot \frac{-3}{\sqrt{26}} + 1 \cdot \frac{1}{\sqrt{26}} = \frac{3}{\sqrt{26}} + \frac{1}{\sqrt{26}} = \frac{4}{\sqrt{26}} \]

### Step 4: Calculate the projection of \(\vec{a}\) onto \(\vec{b}\)
The projection of \(\vec{a}\) onto \(\vec{b}\) is given by:
\[ \text{proj}_{\vec{b}}(\vec{a}) = (\vec{a} \cdot \hat{u}_{\vec{b}}) \hat{u}_{\vec{b}} \]
Substituting the values:
\[ \text{proj}_{\vec{b}}(\vec{a}) = \left(\frac{4}{\sqrt{26}}\right) \left(\frac{4\hat{i} - 3\hat{j} + \hat{k}}{\sqrt{26}}\right) \]
\[ = \frac{4}{26}(4\hat{i} - 3\hat{j} + \hat{k}) = \frac{16}{26}\hat{i} - \frac{12}{26}\hat{j} + \frac{4}{26}\hat{k} \]
\[ = \frac{8}{13}\hat{i} - \frac{6}{13}\hat{j} + \frac{2}{13}\hat{k} \]

### Final Answer
Thus, the projection of \(\vec{a}\) onto \(\vec{b}\) is:
\[ \text{proj}_{\vec{b}}(\vec{a}) = \frac{8}{13}\hat{i} - \frac{6}{13}\hat{j} + \frac{2}{13}\hat{k} \]