Find the vector equation of the line passing | vedclass

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(A) Let the required line be parallel to vector $\vec{b} = b_{1}\hat{i} + b_{2}\hat{j} + b_{3}\hat{k}$.The position vector of the point $(1, 2, 3)$ is

Vedclass · Accepted Answer

$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-3\hat{i} + 5\hat{j} + 4\hat{k})$

Vedclass · Answer

(A) Let the required line be parallel to vector $\vec{b} = b_{1}\hat{i} + b_{2}\hat{j} + b_{3}\hat{k}$.The position vector of the point $(1, 2, 3)$ is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.The equation of the line passing through $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r} = \vec{a} + \lambda\vec{b}$.Since the line is parallel to the given planes,the vector $\vec{b}$ must be perpendicular to the normals of both planes.The normals are $\vec{n}_{1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{n}_{2} = 3\hat{i} + \hat{j} + \hat{k}$.Thus,$\vec{b} = \vec{n}_{1} 	imes \vec{n}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 3 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 + 3) = -3\hat{i} + 5\hat{j} + 4\hat{k}$.Substituting $\vec{a}$ and $\vec{b}$ into the line equation,we get $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-3\hat{i} + 5\hat{j} + 4\hat{k})$.

Find the vector equation of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 5$ and $\vec{r} \cdot (3\hat{i} + \hat{j} + \hat{k}) = 6$.

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