Plane and Line Geometry

00:00 7 marks

Consider the plane \(\Pi_1\) with equation \(2x – y + 3z = 5\) and the line \(L_1\) with equation \(r = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 5 \\ 1 \end{pmatrix}\).

(a) Show that the line \(L_1\) is parallel to the plane \(\Pi_1\). [3]

(b) Find the equation of the plane \(\Pi_2\) which contains the line \(L_1\) and is perpendicular to the plane \(\Pi_1\). Give your answer in the form \(ax + by + cz = d\). [4]

Part a

5 minutes 3 marks

Question

Show that the line \(L_1\) is parallel to the plane \(\Pi_1\). [3]

Hint Hint

Hint

Compute the dot product of the plane’s normal vector \(\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) with the line’s direction vector \(\begin{pmatrix} 1 \\ 5 \\ 1 \end{pmatrix}\) and verify that it equals zero.

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Complete Answer

The dot product is \(2(1) + (-1)(5) + 3(1) = 0\), so \(L_1\) is parallel to \(\Pi_1\).

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Part b

5 minutes 4 marks

Question

Find the equation of the plane \(\Pi_2\) which contains the line \(L_1\) and is perpendicular to the plane \(\Pi_1\). Give your answer in the form \(ax + by + cz = d\). [4]

Hint Hint

Hint

Find a normal vector for \(\Pi_2\) by taking the cross product of \(L_1\)’s direction vector and \(\Pi_1\)’s normal vector. Then substitute a point from \(L_1\) into the plane equation to solve for the constant.

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Complete Answer

The normal vector of \(\Pi_2\) is \(\begin{pmatrix} 16 \\ -1 \\ -11 \end{pmatrix}\) and using the point \((1, 0, -1)\), the plane equation is \(16x – y – 11z = 27\).

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