# IB DP Math AA Topic : AHL 3.17: Use of normal vector to obtain the form r⋅n=a⋅n HL Paper 2

## Question

The points P(−1, 2, − 3), Q(−2, 1, 0), R(0, 5, 1) and S form a parallelogram, where S is diagonally opposite Q.

a.Find the coordinates of S.[2]

b.The vector product $$\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}} { – 13} \\ 7 \\ m \end{array}} \right)$$. Find the value of m .
[2]

c.Hence calculate the area of parallelogram PQRS.[2]

d.Find the Cartesian equation of the plane, $${\prod _1}$$ , containing the parallelogram PQRS.[3]

e.Write down the vector equation of the line through the origin (0, 0, 0) that is perpendicular to the plane $${\prod _1}$$ .[1]

f.Hence find the point on the plane that is closest to the origin.[3]

g.A second plane, $${\prod _2}$$ , has equation x − 2y + z = 3. Calculate the angle between the two planes.[4]

## Markscheme

$$\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}} { – 1} \\ { – 1} \\ 3 \end{array}} \right)$$ , $$\overrightarrow {{\text{SR}}} = \left( {\begin{array}{*{20}{c}} {0 – x} \\ {5 – y} \\ {1 – z} \end{array}} \right)$$     (M1)

point S = (1, 6, −2)     A1

[2 marks]

a.

$$\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}} { – 1} \\ { – 1} \\ 3 \end{array}} \right)$$$$\overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ 1 \end{array}} \right)$$     A1

$$\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}} { – 13} \\ 7 \\ { – 2} \end{array}} \right)$$

m = −2     A1

[2 marks]

b.

area of parallelogram PQRS $$= \left| {\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} } \right| = \sqrt {{{( – 13)}^2} + {7^2} + {{( – 2)}^2}}$$     M1

$$= \sqrt {222} = 14.9$$     A1

[2 marks]

c.

equation of plane is −13x + 7y − 2z = d     M1A1

substituting any of the points given gives d = 33

−13x + 7y − 2z = 33     A1

[3 marks]

d.

equation of line is $$\boldsymbol{r} = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \\ 0 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 13} \\ 7 \\ { – 2} \end{array}} \right)$$     A1

Note: To get the A1 must have $$\boldsymbol{r} =$$ or equivalent.

[1 mark]

e.

$$169\lambda + 49\lambda + 4\lambda = 33$$     M1

$$\lambda = \frac{{33}}{{222}}{\text{ }}( = 0.149…)$$     A1

closest point is $$\left( { – \frac{{143}}{{74}},\frac{{77}}{{74}}, – \frac{{11}}{{37}}} \right){\text{ }}\left( { = ( – 1.93,{\text{ 1.04, – 0.297)}}} \right)$$     A1

[3 marks]

f.

angle between planes is the same as the angle between the normals     (R1)

$$\cos \theta = \frac{{ – 13 \times 1 + 7 \times – 2 – 2 \times 1}}{{\sqrt {222} \times \sqrt 6 }}$$     M1A1

$$\theta = 143^\circ$$ (accept $$\theta = 37.4^\circ$$ or 2.49 radians or 0.652 radians)     A1

[4 marks]

g.

## Question

The points A and B have coordinates (1, 2, 3) and (3, 1, 2) relative to an origin O.

(i)     Find $$\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}}$$ .

(ii)     Determine the area of the triangle OAB.

(iii)     Find the Cartesian equation of the plane OAB.

[5]
a.

(i)     Find the vector equation of the line $${L_1}$$ containing the points A and B.

(ii)     The line $${L_2}$$ has vector equation $$\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ 3 \end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}} 1 \\ 3 \\ 2 \end{array}} \right)$$.

Determine whether or not $${L_1}$$ and $${L_2}$$ are skew.

[7]
b.

## Markscheme

(i)     $$\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} =$$ i + 7j – 5k     A1

(ii)     area $$= \frac{1}{2}|$$i + 7j – 5k$$| = \frac{{5\sqrt 3 }}{2}{\text{(4.33)}}$$     M1A1

(iii)     equation of plane is $$x + 7y – 5z = k$$     M1

$$x + 7y – 5z = 0$$     A1

[5 marks]

a.

(i)     direction of line = (3i + j + 2k) – (i + 2j + 3k) = 2ijk     M1A1

equation of line is

r = (i + 2j + 3k) + $$\lambda$$(2ijk)     A1

(ii)     at a point of intersection,

$$1 + 2\lambda = 2 + \mu$$

$$2 – \lambda = 4 + 3\mu$$     M1A1

$$3 – \lambda = 3 + 2\mu$$

solving the $${2^{{\text{nd}}}}$$ and $${3^{{\text{rd}}}}$$ equations, $$\lambda = 4{\text{, }}\mu = – 2$$     A1

these values do not satisfy the $${1^{{\text{st}}}}$$ equation so the lines are skew     R1

[7 marks]

b.
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