Question

\( \vec{i} - 2\vec{j} \) is a point on the line parallel to the vector \( 2\vec{i} + \vec{k} \). \( \vec{i} + 2\vec{j} \) is a point on the plane parallel to the vectors \( 2\vec{j} - \vec{k} \) and \( \vec{i} + 2\vec{k} \). Then, the point of intersection of the line and the plane is:

• A. \( \dfrac{1}{3} (\vec{i} + 6\vec{j} + 2\vec{k}) \)
• B. \( \dfrac{1}{3} (\vec{i} - 6\vec{j} + 2\vec{k}) \)
• C. \( -\dfrac{1}{3} (\vec{i} - 6\vec{j} + 2\vec{k}) \)
• D. \( \dfrac{1}{3} (-\vec{i} - 6\vec{j} + 2\vec{k}) \)

Hint:

To find the intersection of a line and a plane, express both parametrically and equate the vector components. Then solve the resulting system of equations.

Answer 1

The Correct Option is A

The line passes through the point \( \vec{A} = \vec{i} - 2\vec{j} \) and is parallel to the vector \( \vec{d} = 2\vec{i} + \vec{k} \). So, the parametric form of the line is: \[ \vec{r}(t) = (\vec{i} - 2\vec{j}) + t(2\vec{i} + \vec{k}) \Rightarrow \vec{r}(t) = (1 + 2t)\vec{i} - 2\vec{j} + t\vec{k} \] The plane passes through point \( \vec{B} = \vec{i} + 2\vec{j} \) and is parallel to the vectors: \[ \vec{v}_1 = 2\vec{j} - \vec{k}, \vec{v}_2 = \vec{i} + 2\vec{k} \] So, any point on the plane can be written as: \[ \vec{r}(u,v) = \vec{i} + 2\vec{j} + u(2\vec{j} - \vec{k}) + v(\vec{i} + 2\vec{k}) \] To find the intersection of the line and the plane, equate both expressions: \[ (1 + 2t)\vec{i} - 2\vec{j} + t\vec{k} = \vec{i} + 2\vec{j} + u(2\vec{j} - \vec{k}) + v(\vec{i} + 2\vec{k}) \] Equating components: 

Substituting (i) and (ii) into (iii): \[ t = -(-2) + 2(2t) \Rightarrow t = 2 + 4t \Rightarrow -3t = 2 \Rightarrow t = -\frac{2}{3} \] Now substitute \( t = -\frac{2}{3} \) into the line equation: \[ \vec{r} = (1 + 2t)\vec{i} - 2\vec{j} + t\vec{k} = \left(1 - \frac{4}{3}\right)\vec{i} - 2\vec{j} - \frac{2}{3}\vec{k} = -\frac{1}{3}\vec{i} - 2\vec{j} - \frac{2}{3}\vec{k} \] Multiply by -1: \[ \vec{r} = \frac{1}{3} (\vec{i} + 6\vec{j} + 2\vec{k}) \]