The vector `veca=-hati+2hatj+hatk` is rotated through a right angle, passing through the y-axis in its way and the resulting vector is `vecb`.Then the projection of `3veca+sqrt2b` on `vec c=5hati+4hatj+3hatk` is:

Options

A. `3sqrt2`
B. 1
C. `2sqrt3`
D. sqrt6

Answer

To solve the problem, we will follow these steps:

1. **Define the vectors**:
Given \( \vec{a} = -\hat{i} + 2\hat{j} + \hat{k} \).

2. **Rotation of vector \( \vec{a} \)**:
The vector \( \vec{a} \) is rotated through a right angle about the y-axis to obtain vector \( \vec{b} \). The rotation about the y-axis can be represented using the rotation matrix:
\[ R_y(\theta) = \begin{pmatrix} \cos \theta & 0 & \sin \theta \\ 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta \end{pmatrix} \]
For a right angle rotation (\(\theta = 90^\circ\)), we have:
\[ R_y(90^\circ) = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix} \]
Applying this to \( \vec{a} \):
\[ \vec{b} = R_y(90^\circ) \cdot \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \cdot -1 + 0 \cdot 2 + 1 \cdot 1 \\ 0 \cdot -1 + 1 \cdot 2 + 0 \cdot 1 \\ -1 \cdot -1 + 0 \cdot 2 + 0 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \]
Thus, \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \).

3. **Projection of \( 3\vec{a} + \sqrt{2}\vec{b} \)**:
We first calculate \( 3\vec{a} + \sqrt{2}\vec{b} \):
\[ 3\vec{a} = 3(-\hat{i} + 2\hat{j} + \hat{k}) = -3\hat{i} + 6\hat{j} + 3\hat{k} \]
\[ \sqrt{2}\vec{b} = \sqrt{2}(\hat{i} + 2\hat{j} - \hat{k}) = \sqrt{2}\hat{i} + 2\sqrt{2}\hat{j} - \sqrt{2}\hat{k} \]
Now, adding these two vectors:
\[ 3\vec{a} + \sqrt{2}\vec{b} = (-3 + \sqrt{2})\hat{i} + (6 + 2\sqrt{2})\hat{j} + (3 - \sqrt{2})\hat{k} \]

4. **Define vector \( \vec{c} \)**:
Given \( \vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k} \).

5. **Calculate the projection**:
The projection of vector \( \vec{v} \) onto vector \( \vec{c} \) is given by:
\[ \text{proj}_{\vec{c}} \vec{v} = \frac{\vec{v} \cdot \vec{c}}{|\vec{c}|^2} \vec{c} \]
First, we need to calculate \( \vec{v} \cdot \vec{c} \):
\[ \vec{v} = (-3 + \sqrt{2})\hat{i} + (6 + 2\sqrt{2})\hat{j} + (3 - \sqrt{2})\hat{k} \]
\[ \vec{v} \cdot \vec{c} = (-3 + \sqrt{2}) \cdot 5 + (6 + 2\sqrt{2}) \cdot 4 + (3 - \sqrt{2}) \cdot 3 \]
Simplifying this:
\[ = -15 + 5\sqrt{2} + 24 + 8\sqrt{2} + 9 - 3\sqrt{2} = 18 + 10\sqrt{2} \]

6. **Calculate \( |\vec{c}|^2 \)**:
\[ |\vec{c}|^2 = 5^2 + 4^2 + 3^2 = 25 + 16 + 9 = 50 \]

7. **Final projection calculation**:
\[ \text{proj}_{\vec{c}} \vec{v} = \frac{18 + 10\sqrt{2}}{50} \vec{c} \]

8. **Final answer**:
The projection of \( 3\vec{a} + \sqrt{2}\vec{b} \) on \( \vec{c} \) is:
\[ \frac{(18 + 10\sqrt{2})}{50} (5\hat{i} + 4\hat{j} + 3\hat{k}) \]