ONE EIGHTH CONTEST 2025
Ghana National College - 65 points
Tema Secondary School - 33 points
Effiduase SHCS - 28 points
ONE EIGHTH CONTEST 2025
Mankranso SHS - 43 points
St Paul's SHS, Denu - 23 points
Adonten SHS - 14 points
ONE EIGHTH CONTEST 2025
St. Thomas Aquinas SHS - 51 points
Tamale SHS - 37 points
O'Reilly SHS - 25 points
ROUND 1
PREAMBLE
Find the scalar product of the vectors A plus B and A minus B given,
FIRST QUESTION
$A=2i+3j$ and $B=i-4j$
$ANSWER = -4$
SOLUTION 1
$(A+B).(A-B)$ = $|A|^2 - |B|^2$
$|A|^2$ = $(2)^2 + (3)^2$ = $4 + 9$ = 13
$|B|^2$ = $(1)^2 + (-4)^2$ = $1 + 16$ = 17
$|A|^2 - |B|^2$ = 13 - 17
$= -4$
SOLUTION 2
$A+B=2i+3j+i-4j$
$=3i-j$
$A-B=2i+3j-(i-4j)$
$=2i-i+3j+4j$
$=i+7j$
$(A+B).(A-B)=(3i-j).(i+7j)$
$=(3 \times 1)+(-1 \times 7)$
$=3-7$
$=-4$
$ANSWER = -4$
SECOND QUESTION
$A=3i+4j$ and $B=5i-12j$
$ANSWER = -144$
SOLUTION
$(A+B).(A-B)$ = $|A|^2 - |B|^2$
$|A|^2$ = $(3)^2 + (4)^2$ = $9 + 16$ = 25
$|B|^2$ = $(5)^2 + (-12)^2$ = $25 + 144$ = 169
$|A|^2 - |B|^2$ = 25 - 169
$= -144$
THIRD QUESTION
$A=2i-5j$ and $B=4i-3j$
$ANSWER = 4$
SOLUTION
$(A+B).(A-B)$ = $|A|^2 - |B|^2$
$|A|^2$ = $(2)^2 + (-5)^2$ = $4 + 25$ = 29
$|B|^2$ = $(4)^2 + (-3)^2$ = $16 + 9$ = 25
$|A|^2 - |B|^2$ = 29 - 25
$= 4$
PRACTICE QUESTIONS
Find the scalar product of the vectors A plus B and A minus B.
1. $A=4\mathbf{i}+3\mathbf{j}$ and $B=\mathbf{i}+2\mathbf{j}$. Find $(A+B)\cdot(A-B)$.
ANSWER: $20$
SOLUTION 1
$(A+B)\cdot(A-B)=|A|^2-|B|^2$
$|A|^2=4^2+3^2=25$
$|B|^2=1^2+2^2=5$
$|A|^2-|B|^2=25-5$
$=20$
SOLUTION 2
$A+B=5\mathbf{i}+5\mathbf{j}$
$A-B=3\mathbf{i}+\mathbf{j}$
$(A+B)\cdot(A-B)=(5)(3)+(5)(1)$
$=15+5$
$=20$
2. $A=2\mathbf{i}-\mathbf{j}$ and $B=3\mathbf{i}+4\mathbf{j}$. Find $(A+B)\cdot(A-B)$.
ANSWER: $-20$
SOLUTION 1
$(A+B)\cdot(A-B)=|A|^2-|B|^2$
$|A|^2=2^2+(-1)^2=5$
$|B|^2=3^2+4^2=25$
$|A|^2-|B|^2=5-25$
$=-20$
SOLUTION 2
$A+B=5\mathbf{i}+3\mathbf{j}$
$A-B=-\mathbf{i}-5\mathbf{j}$
$(A+B)\cdot(A-B)=(5)(-1)+(3)(-5)$
$=-5-15$
$=-20$
3. $A=5\mathbf{i}+2\mathbf{j}$ and $B=\mathbf{i}-3\mathbf{j}$. Find $(A+B)\cdot(A-B)$.
ANSWER: $19$
SOLUTION 1
$(A+B)\cdot(A-B)=|A|^2-|B|^2$
$|A|^2=5^2+2^2=29$
$|B|^2=1^2+(-3)^2=10$
$|A|^2-|B|^2=29-10$
$=19$
SOLUTION 2
$A+B=6\mathbf{i}-\mathbf{j}$
$A-B=4\mathbf{i}+5\mathbf{j}$
$(A+B)\cdot(A-B)=(6)(4)+(-1)(5)$
$=24-5$
$=19$
4. $A=\mathbf{i}+6\mathbf{j}$ and $B=4\mathbf{i}-2\mathbf{j}$. Find $(A+B)\cdot(A-B)$.
ANSWER: $17$
SOLUTION 1
$(A+B)\cdot(A-B)=|A|^2-|B|^2$
$|A|^2=1^2+6^2=37$
$|B|^2=4^2+(-2)^2=20$
$|A|^2-|B|^2=37-20$
$=17$
SOLUTION 2
$A+B=5\mathbf{i}+4\mathbf{j}$
$A-B=-3\mathbf{i}+8\mathbf{j}$
$(A+B)\cdot(A-B)=(5)(-3)+(4)(8)$
$=-15+32$
$=17$
5. $A=7\mathbf{i}+\mathbf{j}$ and $B=2\mathbf{i}+4\mathbf{j}$. Find $(A+B)\cdot(A-B)$.
ANSWER: $30$
SOLUTION 1
$(A+B)\cdot(A-B)=|A|^2-|B|^2$
$|A|^2=7^2+1^2=50$
$|B|^2=2^2+4^2=20$
$|A|^2-|B|^2=50-20$
$=30$
SOLUTION 2
$A+B=9\mathbf{i}+5\mathbf{j}$
$A-B=5\mathbf{i}-3\mathbf{j}$
$(A+B)\cdot(A-B)=(9)(5)+(5)(-3)$
$=45-15$
$=30$
6. $A=-3\mathbf{i}+6\mathbf{j}$ and $B=5\mathbf{i}+2\mathbf{j}$. Find $(A+B)\cdot(A-B)$.
ANSWER: $16$
SOLUTION 1
$(A+B)\cdot(A-B)=|A|^2-|B|^2$
$|A|^2=(-3)^2+6^2=45$
$|B|^2=5^2+2^2=29$
$|A|^2-|B|^2=45-29$
$=16$
SOLUTION 2
$A+B=2\mathbf{i}+8\mathbf{j}$
$A-B=-8\mathbf{i}+4\mathbf{j}$
$(A+B)\cdot(A-B)=(2)(-8)+(8)(4)$
$=-16+32$
$=16$