Question 2:
2 | (a) |  4   3 
   
−5 . 6 −15
   
   
 1  −2
 3 
 
6
 
 
−2
or
3
2 2 2
� 6 � = √3 +6 +2
−2
4 3
�−5�.� 6 � = 4×3+(−5)×6+
5 1 −2
1 ×(−2) | M1
M1
A1
[3] | 1.1a
1.1
1.1 | Substitution into formula.
Condone missing mod signs on
top
Soi | Numerator can be
3
where q is vector f�r𝐪𝐪om.� C6 to� �point
on Π −2
 4   3 
   
Or substituting r= −5 +λ 6
   
   
 1  −2
into equation of Π to find a value
for λ.
(b) | 5 4 1
𝒓𝒓= �2�−�3� = �−1�
4 1 3 | M1 | Finding the vector r from a point
on one line to a point on the other
1 1 | *M1 | Using dot product to find an
expression for cosθ | NB Ignore attempts to use the
formula for the distance between
2 skew lines.
�−1�.�−2� = √11√6cosθ
3 1
6 | A1 | can be implied by ,
or or 2.402 or 42.4º
or 137.6 º cos𝑐𝑐 = 0.739
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = ±
√66 | dep*M1 | 𝑐𝑐 = 0.740
5 2 | A1
Finding the vector r from a point
on one line to a point on the other
Using dot product to find an
expression for cosθ
Alternate method 1
5  1  4
     
r= 2 +µ −2 − 3
     
     
4  1  1 | M1 | 3.1a | 3.1a | Finding the vector from a
particular point on one line to a
general point on the other | Finding the vector from a
particular point on one line to a
general point on the other
5  1  4  1 
       
 2  +µ  −2  −  3 .  −2  =0
       
4  1  1  1  | *M1 | 1.1a | Dotting the vector found with one
of the direction vectors and
setting to 0
(μ = -1 so)
4
�4� | (μ = -1 so)
4 | A1 | 1.1 | Solving to find (a parameter | or ( and)
1 5
𝜆𝜆 = −2 �1�
value and hence) position vector
or coordinates of equivalent point
on other line
3
d = (4−4)2 +(3−4)2 +(1−3)2 | dep*M1 | 1.1 | Distance formula for their 2 | 2
points
5 | A1 | 1.1
Alternate method 2
5  1  4
     
r= 2 +µ −2 − 3
     
     
4  1  1 | M1 | Finding the vector from a
particular point on one line to a
general point on the other | As above
*M1
A1
2
|𝒓𝒓|= �6𝜇𝜇 +12 𝜇𝜇 +11 | dep*M1 | Attempts to find parameter
(Calculus or completes square)
𝜇𝜇 = −1 | A1
Alternate method 3
𝑎𝑎 = √5
5 4 1
𝒓𝒓= �2�−�3� = �−1� | 5 4 1 | M1 | Finding the vector r from a point
on one line to a point on the other
Dotting the vector found with one
of the direction vectors and
setting to 0
Finding the vector from a
particular point on one line to a
general point on the other
Finding the vector r from a point
on one line to a point on the other
1 1 5 | *M1 | Calculates cross product of ‘their
r’ and direction vector. | Allow if full method seen or if 2
terms correct
� … �−1�×�−2�� = � 2 � | A1
3 1 −1
2 2 2
= √5 +2 +1 = √30
√30
𝑎𝑎 = = √5
1 | dep*M1
A1
��−2�� | [5]