With reference to the origin O, the points N, P and Q have position vectors n\mathbf{n}, p\mathbf{p} and q\mathbf{q} respectively, and Π\Pi has an equation r⋅n=0\mathbf{r}\cdot\mathbf{n}=0. It is known that p⋅n=q⋅n≠0\mathbf{p}\cdot\mathbf{n}=\mathbf{q}\cdot\mathbf{n}\ne 0.

  1. ((i))

    Show that PQ→\overrightarrow{PQ} is perpendicular to n\mathbf{n}. Hence describe the geometrical relationship between the line PQ and the plane Π\Pi.

  2. ((ii))

    Find a vector equation of the plane that contains points P and Q, and is perpendicular to plane Π\Pi, leaving your answer in terms of n\mathbf{n}, p\mathbf{p} and q\mathbf{q}.

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Worked solution

5(b)((i))

1. Express the direction vector of PQ and compute its scalar product with $\mathbf{n}$.

PQ→⋅n=(q−p)⋅n\overrightarrow{PQ} \cdot \mathbf{n} = (\mathbf{q} - \mathbf{p}) \cdot \mathbf{n}
=q⋅n−p⋅n= \mathbf{q}\cdot\mathbf{n} - \mathbf{p}\cdot\mathbf{n}

2. Show that the scalar product is zero and conclude perpendicularity.

Since it is given that p⋅n=q⋅n\mathbf{p}\cdot\mathbf{n} = \mathbf{q}\cdot\mathbf{n},
PQ→⋅n=q⋅n−q⋅n=0\overrightarrow{PQ} \cdot \mathbf{n} = \mathbf{q}\cdot\mathbf{n} - \mathbf{q}\cdot\mathbf{n} = 0
Since the scalar product is 00, PQ→\overrightarrow{PQ} is perpendicular to n\mathbf{n} (shown).

3. Deduce the geometrical relationship between line PQ and plane $\Pi$.

The normal vector to plane Π\Pi is n\mathbf{n}.
Because the direction vector of line PQ (PQ→\overrightarrow{PQ}) is perpendicular to the normal of Π\Pi, line PQ is parallel to plane Π\Pi.
For P to lie on Π\Pi, we must have p⋅n=0\mathbf{p}\cdot\mathbf{n} = 0. However, we are given p⋅n≠0\mathbf{p}\cdot\mathbf{n} \ne 0, so P does not lie on Π\Pi.
Hence, the line PQ is strictly parallel to plane Π\Pi and does not lie on it.

Answer

Line PQ is parallel to plane Π\Pi (and does not lie on it).

5(b)((ii))

1. Identify two non-parallel direction vectors parallel to the required plane.

The new plane is perpendicular to Π\Pi, so the normal vector of Π\Pi, which is n\mathbf{n}, is a direction vector parallel to the new plane.

Since the plane contains points PP and QQ, the vector PQ→=q−p\overrightarrow{PQ} = \mathbf{q} - \mathbf{p} is another direction vector parallel to the plane.

(From part (i), n\mathbf{n} and q−p\mathbf{q} - \mathbf{p} are perpendicular, so they are non-parallel and form a valid pair of direction vectors for a plane).

2. Form the parametric vector equation of the plane.

Using the position vector of point PP (which is p\mathbf{p}) as a known point on the plane, the vector equation of the plane is given by:

r=p+λ(q−p)+μn\mathbf{r} = \mathbf{p} + \lambda(\mathbf{q} - \mathbf{p}) + \mu\mathbf{n}
where λ,μ∈R\lambda, \mu \in \mathbb{R}.

Answer

r=p+λ(q−p)+μn,where λ,μ∈R\mathbf{r} = \mathbf{p} + \lambda(\mathbf{q} - \mathbf{p}) + \mu\mathbf{n}, \quad \text{where } \lambda, \mu \in \mathbb{R}

AI analysis — may contain mistakes.

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