Find the vector equation of the line through the origin, which is perpendicular to the plane `vec(r) . (hati - 2 hatj + hatk)` = 3 .

Answer

To find the vector equation of the line through the origin that is perpendicular to the given plane, we can follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane is given as:
\[ \vec{r} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 3 \]
The normal vector of the plane can be directly taken from the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) in the equation. Thus, the normal vector \(\vec{n}\) is:
\[ \vec{n} = \hat{i} - 2\hat{j} + \hat{k} \]

### Step 2: Determine the direction vector of the line
Since the line is perpendicular to the plane, its direction vector \(\vec{d}\) will be the same as the normal vector of the plane:
\[ \vec{d} = \hat{i} - 2\hat{j} + \hat{k} \]

### Step 3: Write the vector equation of the line
The vector equation of a line can be expressed in the form:
\[ \vec{r} = \vec{a} + \lambda \vec{b} \]
where \(\vec{a}\) is a point on the line (which is the origin in this case, \(\vec{0}\)), \(\lambda\) is a scalar parameter, and \(\vec{b}\) is the direction vector. Here, \(\vec{a} = \vec{0}\) and \(\vec{b} = \vec{d}\).

Thus, substituting the values, we get:
\[ \vec{r} = \vec{0} + \lambda (\hat{i} - 2\hat{j} + \hat{k}) \]
This simplifies to:
\[ \vec{r} = \lambda (\hat{i} - 2\hat{j} + \hat{k}) \]

### Final Answer
The vector equation of the line through the origin and perpendicular to the plane is:
\[ \vec{r} = \lambda (\hat{i} - 2\hat{j} + \hat{k}) \]

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