Find the vector equation of the line passing | vedclass

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Question

(A) The position vector of the point $(1, 2, 3)$ is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.The plane is given by $\vec{r} \cdot (\hat{i} + 2\hat{j}

Vedclass · Accepted Answer

$\vec{l} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + 2\hat{j} - 5\hat{k})$

Vedclass · Answer

(A) The position vector of the point $(1, 2, 3)$ is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.The plane is given by $\vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0$. The normal vector to the plane is $\vec{n} = \hat{i} + 2\hat{j} - 5\hat{k}$.$A$ line perpendicular to the plane will have the same direction as the normal vector of the plane. Thus,the direction vector of the line is $\vec{v} = \hat{i} + 2\hat{j} - 5\hat{k}$.The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to vector $\vec{v}$ is $\vec{r} = \vec{a} + \lambda\vec{v}$.Substituting the values,we get $\vec{l} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + 2\hat{j} - 5\hat{k})$.

Find the vector equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $\vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0$.

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