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2017 A-Level · Paper 1 · Question 6

2017 H2 Maths Paper 1, Question 6

Vectors 8 marks geometrical interpretation of equations
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Part (i)

2 marks

r=a+tb\mathbf{r} = \mathbf{a} + t\mathbf{b} is the vector equation of a straight line: it passes through the point with position vector a\mathbf{a}, and is parallel to (has direction) b\mathbf{b}. As the parameter tt varies over all real numbers, r\mathbf{r} traces out every point on this line.

Part (ii)

3 marks

r⋅n=d\mathbf{r} \cdot \mathbf{n} = d is the vector equation of a plane with unit normal vector n\mathbf{n}: it consists of all points whose position vector r\mathbf{r} has the same projection onto n\mathbf{n}, namely dd.

Since n\mathbf{n} is a unit vector, d=r⋅nd = \mathbf{r} \cdot \mathbf{n} is the length of the projection of r\mathbf{r} onto n\mathbf{n} for any point on the plane, i.e. dd is the perpendicular distance from the origin to the plane.

Part (iii)

3 marks

Approach

The point we want lies on both the line and the plane, so we substitute the line’s equation for r\mathbf{r} into the plane’s equation and solve for tt.

Substituting r=a+tb\mathbf{r} = \mathbf{a} + t\mathbf{b} into r⋅n=d\mathbf{r} \cdot \mathbf{n} = d:

(a+tb)⋅n=d\left( \mathbf{a} + t\mathbf{b} \right) \cdot \mathbf{n} = d
a⋅n+t(b⋅n)=d\mathbf{a} \cdot \mathbf{n} + t\left( \mathbf{b} \cdot \mathbf{n} \right) = d

Since b⋅n≠0\mathbf{b} \cdot \mathbf{n} \neq 0, we may divide by it:

t=d−a⋅nb⋅nt = \frac{d - \mathbf{a} \cdot \mathbf{n}}{\mathbf{b} \cdot \mathbf{n}}

Substituting back into r=a+tb\mathbf{r} = \mathbf{a} + t\mathbf{b}:

r=a+(d−a⋅nb⋅n)b\mathbf{r} = \mathbf{a} + \left( \frac{d - \mathbf{a} \cdot \mathbf{n}}{\mathbf{b} \cdot \mathbf{n}} \right)\mathbf{b}

Since b⋅n≠0\mathbf{b} \cdot \mathbf{n} \neq 0, the line is not parallel to the plane, so it meets the plane at exactly one point.

∴\therefore r=a+(d−a⋅nb⋅n)b\mathbf{r} = \mathbf{a} + \left( \frac{d - \mathbf{a} \cdot \mathbf{n}}{\mathbf{b} \cdot \mathbf{n}} \right)\mathbf{b}, the position vector of that unique point of intersection.

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