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Find the vector equation of the line parallel to the vector  3\hat{i}-2 \hat{j}+3\hat{k} and which passes through the point (1, -2, 3).

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Given, vector = 3\hat{i}-2\hat{j}+6\hat{k}

Point = (1, -2, 3)

We can write this point in vector form as \hat{i}-2\hat{j}+3\hat{k}

Let ,



We must find the vector equation of the line parallel to the vector \overrightarrow{b} and passing through the point 

We know, equation of \overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b} a line passing through a point and parallel to a given vector is denoted as

Where, \lambda \epsilon \mathbb{R}

In other words, we need to find  \overrightarrow{r}

This can be achieved by substituting the values of the vectors in the above equation. We get

\overrightarrow{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )

\Rightarrow \overrightarrow{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )

This can be further rearranged, upon which we get:

\Rightarrow \vec{r}=\hat{i}-2\hat{j}+3\hat{k}+3\lambda \hat{j}+3\hat{k}+6\lambda\hat{k}

\Rightarrow \vec{r}=\hat{i}+3\lambda \hat{i}-2\hat{j}-2\lambda\hat{j}+3\hat{k}+6\lambda\hat{k}

\Rightarrow \vec{r}=\left ( 1-3\lambda \right )\hat{i}+\left ( -2-2\lambda \right )\hat{j}+\left ( 3+6\lambda \right )\hat{k}

Thus, the require vector equation of line is   \vec{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda\left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )

which can also be written as  \left ( 1-3\lambda \right )i+\left ( -2-2 \lambda \right )\hat{j}+\left ( 3+6\lambda \right )\hat{k}

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