$\vec{a},\vec{b},\vec{c}$ are three orthogonal vectors. Given that $\vec{a}=\hat{i}+2\hat{j}+5\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$, the vector $\vec{c}$ is parallel to

The problem asks us to find a vector $\vec{c}$ that is orthogonal to two given vectors, $\vec{a}$ and $\vec{b}$.

Understanding Orthogonal Vectors

Two vectors are considered orthogonal if their dot product is zero. This means they are perpendicular to each other. If a vector $\vec{c}$ is orthogonal to both $\vec{a}$ and $\vec{b}$, it lies in the direction perpendicular to the plane containing $\vec{a}$ and $\vec{b}$.

Given Vectors

We are given:

- $\vec{a}=\hat{i}+2\hat{j}+5\hat{k}$
- $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$

We need to find $\vec{c}$ such that $\vec{c}$ is orthogonal to both $\vec{a}$ and $\vec{b}$.

**Verifying Orthogonality of $\vec{a}$ and $\vec{b}$ **

First, let's check if the given vectors $\vec{a}$ and $\vec{b}$ are indeed orthogonal to each other, as stated in the problem. We calculate their dot product:

$\vec{a}\cdot \vec{b}=(\hat{i}+2\hat{j}+5\hat{k})\cdot (\hat{i}+2\hat{j}-\hat{k})$

$\vec{a}\cdot \vec{b}=(1)(1)+(2)(2)+(5)(-1)$

$\vec{a}\cdot \vec{b}=1+4-5$

$\vec{a}\cdot \vec{b}=0$

Since the dot product is zero, $\vec{a}$ and $\vec{b}$ are orthogonal.

**Finding the Direction of Vector $\vec{c}$ **

A vector that is orthogonal to two other vectors, $\vec{a}$ and $\vec{b}$, is found by calculating their cross product, $\vec{a}\times \vec{b}$. The vector $\vec{c}$ must be parallel to this cross product.

**Calculating the Cross Product $\vec{a}\times \vec{b}$ **

We compute the cross product using the determinant method:

$$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 5\\ 1& 2& -1\end{array}\right|$$

Expanding the determinant:

$$\vec{a}\times \vec{b}=\hat{i}\left|\begin{array}{cc}2& 5\\ 2& -1\end{array}\right|-\hat{j}\left|\begin{array}{cc}1& 5\\ 1& -1\end{array}\right|+\hat{k}\left|\begin{array}{cc}1& 2\\ 1& 2\end{array}\right|$$

$$\vec{a}\times \vec{b}=\hat{i}((2)(-1)-(5)(2))-\hat{j}((1)(-1)-(5)(1))+\hat{k}((1)(2)-(2)(1))$$

$$\vec{a}\times \vec{b}=\hat{i}(-2-10)-\hat{j}(-1-5)+\hat{k}(2-2)$$

$$\vec{a}\times \vec{b}=\hat{i}(-12)-\hat{j}(-6)+\hat{k}(0)$$

$$\vec{a}\times \vec{b}=-12\hat{i}+6\hat{j}$$

Identifying the Parallel Vector

The vector $\vec{c}$ must be parallel to $-12\hat{i}+6\hat{j}$. This means $\vec{c}$ can be written as $k(-12\hat{i}+6\hat{j})$ for some non-zero scalar $k$. We can simplify the direction vector by dividing by a common factor. Dividing by -6, we get:

$$\frac{-12\hat{i}+6\hat{j}}{-6}=2\hat{i}-\hat{j}$$

So, $\vec{c}$ must be parallel to $2\hat{i}-\hat{j}$.

Comparing with Options

Now, we compare this direction vector with the given options:

- Option 1: $\hat{i}+2\hat{j}+3\hat{k}$
- Option 2: $2\hat{i}+\hat{j}$
- Option 3: $2\hat{i}-\hat{j}$
- Option 4: $4\hat{k}$

The vector $2\hat{i}-\hat{j}$ matches Option 3.

Conclusion

The vector $\vec{c}$ must be parallel to the cross product of $\vec{a}$ and $\vec{b}$, which we found to be in the direction of $2\hat{i}-\hat{j}$. Therefore, the vector $\vec{c}$ is parallel to $2\hat{i}-\hat{j}$.