Challenging +1.2 This is a Further Maths question requiring computation of a cross product, then using the perpendicularity condition (dot product = 0) to form a quadratic equation in p. While it involves multiple steps and algebraic manipulation, the procedure is methodical and follows standard techniques without requiring novel insight. The cross product calculation is routine for FP2 students, making this moderately above average difficulty.
Vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\), are given by \(\mathbf{a} = \mathbf{i} + (1-p)\mathbf{j} + (p+2)\mathbf{k}\), \(\mathbf{b} = 2\mathbf{i} + \mathbf{j} + \mathbf{k}\) and \(\mathbf{c} = \mathbf{i} + 14\mathbf{j} + (p-3)\mathbf{k}\) where \(p\) is a constant.
You are given that \(\mathbf{a} \times \mathbf{b}\) is perpendicular to \(\mathbf{c}\).
Determine the possible values of \(p\). [6]
= 1 − ( 1 − 2 p ) + 1 4 ( 2 p + 3 ) + ( p − 3 ) ( 2 p − 1 )
c perpendicular to ab =>c.(ab) = 0 soi
− 1 − 2 p + 2 8 p + 4 2 + 2 p 2 − 7 p + 3 = 0
2 p 2 + 1 9 p + 4 4 = 0
11
p=−4 or p=− cao
Answer
Marks
2
M1*
A1
M1
B1
depM1*
Answer
Marks
A1
3.1a
1.1
1.1
3.1a
1.1
Answer
Marks
1.1
Forming the cross-product. Two
components correct (possibly
unsimplified) or all correct but
for wrong sign globally.
Correctly forming the scalar
product of c and their ab (even
if one or both incorrect)
Rule for two vectors being
perpendicular soi.
Rearranging to 3-term quadratic
equation in p.
Answer
Marks
(p + 4)(2p + 11) = 0
i j k
or expanding 1 1− p p+2
2 1 1
If M0M0 then SCB1 for correctly
forming scalar product of any two
vectors containing p.
“= 0” can be implied by correct
solution.
[6]
Question 5:
5 | 1 2
a × b = 1 − p × 1
p + 2 1
(1 − p ) − 1 ( p + 2 ) 1
= ( p + 2 ) 2 − 1 1
1 1 − (1 − p ) 2
− 1 − 2 p
= 2 p + 3
2 p − 1
1 − 1 − 2 p
1 4 . 2 p + 3
p − 3 2 p − 1
= 1 − ( 1 − 2 p ) + 1 4 ( 2 p + 3 ) + ( p − 3 ) ( 2 p − 1 )
c perpendicular to ab =>c.(ab) = 0 soi
− 1 − 2 p + 2 8 p + 4 2 + 2 p 2 − 7 p + 3 = 0
2 p 2 + 1 9 p + 4 4 = 0
11
p=−4 or p=− cao
2 | M1*
A1
M1
B1
depM1*
A1 | 3.1a
1.1
1.1
3.1a
1.1
1.1 | Forming the cross-product. Two
components correct (possibly
unsimplified) or all correct but
for wrong sign globally.
Correctly forming the scalar
product of c and their ab (even
if one or both incorrect)
Rule for two vectors being
perpendicular soi.
Rearranging to 3-term quadratic
equation in p.
(p + 4)(2p + 11) = 0 | i j k
or expanding 1 1− p p+2
2 1 1
If M0M0 then SCB1 for correctly
forming scalar product of any two
vectors containing p.
“= 0” can be implied by correct
solution.
[6]
Vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$, are given by $\mathbf{a} = \mathbf{i} + (1-p)\mathbf{j} + (p+2)\mathbf{k}$, $\mathbf{b} = 2\mathbf{i} + \mathbf{j} + \mathbf{k}$ and $\mathbf{c} = \mathbf{i} + 14\mathbf{j} + (p-3)\mathbf{k}$ where $p$ is a constant.
You are given that $\mathbf{a} \times \mathbf{b}$ is perpendicular to $\mathbf{c}$.
Determine the possible values of $p$. [6]
\hfill \mbox{\textit{OCR Further Pure Core 2 2024 Q5 [6]}}