M5 June 2017 Q3

EdexcelOld spec15 marksMoments Vectors

3. The position vectors of the points \(P\) and \(Q\) on a rigid body are \((\mathbf{i} - 2\mathbf{j} + 3\mathbf{k})\) m and \((\mathbf{i} - \mathbf{j} + \mathbf{k})\) m respectively, relative to a fixed origin \(O\). A force \(\mathbf{F}_1\) of magnitude 6 N acts at \(P\) in the direction \((\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})\). A force \(\mathbf{F}_2\) of magnitude 14 N acts at \(Q\) in the direction \((3\mathbf{i} - 6\mathbf{j} + 2\mathbf{k})\). When a force \(\mathbf{F}_3\) acts at \(O\), which is also a point on the rigid body, the system of three forces is equivalent to a couple of moment \(\mathbf{G}\)

(a) Find \(\mathbf{F}_3\) (5)

(b) Find \(\mathbf{G}\) (3)

When an additional force \(\mathbf{F}_4 = (\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})\) N also acts at \(O\), the system of four forces is equivalent to a single force \(\mathbf{R}\).

(d) Find an equation of the line of action of \(\mathbf{R}\) in the form \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\), where \(\mathbf{a}\) and \(\mathbf{b}\) are constant vectors and \(t\) is a parameter. (6)

Mark scheme (a)

Scheme	Marks
\(\mathbf{F}_1 = \tfrac{1}{3}(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \times 6 = (2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k})\) N	M1 A1
\(\mathbf{F}_2 = \tfrac{1}{7}(3\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}) \times 14 = (6\mathbf{i} - 12\mathbf{j} + 4\mathbf{k})\) N	A1
\(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0} \Rightarrow \mathbf{F}_3 = (-8\mathbf{i} + 16\mathbf{j} - 8\mathbf{k})\) N	M1 A1
	(5)

Notes

First M1 for a complete method to find either \(\mathbf{F}_1\) or \(\mathbf{F}_2\)

First A1 for \(\mathbf{F}_1\)

Second A1 for \(\mathbf{F}_2\)

Second M1 for equating the sum of the 3 forces to zero and solving for \(\mathbf{F}_3\)

Third A1 for a correct \(\mathbf{F}_3\)

Mark scheme (b)

Scheme	Marks
\(\mathbf{G} = (\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) \times (2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}) + (\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (6\mathbf{i} - 12\mathbf{j} + 4\mathbf{k})\)	M1
\(= (4\mathbf{i} + 2\mathbf{j}) + (8\mathbf{i} + 2\mathbf{j} - 6\mathbf{k})\)	A1 either
\(= (12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})\) Nm	A1
	(3)

Notes

First M1 for taking moments about \(O\) (or possibly another point)

Allow \(\mathbf{F} \times \mathbf{r}\)

First A1 for a correct cross-product (either)

Second A1 for \((12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})\)

Scheme	Marks
\(\mathbf{R} = (\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})\) N	B1
	(1)

Mark scheme (d)

Scheme	Marks
\((x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \times (\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) = (12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})\)	M1 A1
\((4y - 3z)\mathbf{i} + (z - 4x)\mathbf{j} + (3x - y)\mathbf{k} = (12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})\)	A1
One solution is \(x = -1,\ y = 3,\ z = 0\)	B1
\(\mathbf{r} = (-\mathbf{i} + 3\mathbf{j}) + t(\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})\)	M1 A1
	(6)
	(15 marks)

Notes

First M1 for taking moments about \(O\) (or possibly another point)

\((x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \times\) their \(\mathbf{R}\) = their \(\mathbf{G}\) (not evaluated)

First A1 for a correct equation (cross products not evaluated)

Second A1 for 3 correct sim equations.

B1 for a correct point on their line (this may appear in their equation)

Second M1 for \(\mathbf{r} = \mathbf{a} + t(\text{their } \mathbf{R})\)

Second A1 for \(\mathbf{r} = (-\mathbf{i} + 3\mathbf{j}) + t(\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})\) or any other correct answer

N.B. Need \(\mathbf{r} = \ldots\)

(Corrected from the printed mark scheme: the second line is printed as \(((4y - 3z)\mathbf{i} + (z - 4x)\mathbf{j} + (3x - y)6\mathbf{k} = (12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})\), with a stray 6 and bracket.)