1. O is the origin, \(\vec{OP} = p\) and \(\vec{OQ} = q\). QT : TP = 2 : 1. Find the position vector of T. Give your answer in terms of p and q, in its simplest form. 2. Without us... 1. O is the origin, \(\vec{OP} = p\) and \(\vec{OQ} = q\). QT : TP = 2 : 1. Find the position vector of T. Give your answer in terms of p and q, in its simplest form. 2. Without using a calculator, work out \(\frac{2}{3} \div 1\frac{1}{5}\). You must show all your working and give your answer as a fraction in its simplest form. 3. (a) The length of the side of a square is 12 cm, correct to the nearest centimetre. Calculate the upper bound for the perimeter of the square. (b) Jo measures the length of a rope and records her measurement correct to the nearest ten centimetres. The upper bound for her measurement is 12.35 m. Write down the measurement she records. Read more

Steps to Solve

1. Find $\vec{QP}$

To find the vector $\vec{QP}$, we use the fact that $\vec{QP} = \vec{OP} - \vec{OQ}$. Since $\vec{OP} = p$ and $\vec{OQ} = q$, we have: $$ \vec{QP} = p - q $$

2. Find $\vec{QT}$

Given the ratio $QT : TP = 2:1$, it means $QT = \frac{2}{3} QP$. Thus, $$ \vec{QT} = \frac{2}{3} \vec{QP} = \frac{2}{3}(p - q) $$

3. Find $\vec{OT}$

The position vector of T, $\vec{OT}$, can be found using $\vec{OT} = \vec{OQ} + \vec{QT}$. We know that $\vec{OQ} = q$ and $\vec{QT} = \frac{2}{3}(p - q)$. Therefore, $$ \vec{OT} = q + \frac{2}{3}(p - q) = q + \frac{2}{3}p - \frac{2}{3}q $$

4. Simplify $\vec{OT}$

Combine the $q$ terms: $$ \vec{OT} = \frac{2}{3}p + q - \frac{2}{3}q = \frac{2}{3}p + \frac{1}{3}q $$

5. Evaluate $\frac{2}{3} \div 1\frac{1}{5}$

First, convert the mixed number $1\frac{1}{5}$ to an improper fraction: $$ 1\frac{1}{5} = \frac{1 \times 5 + 1}{5} = \frac{6}{5} $$ Now, rewrite the division as a multiplication by the reciprocal: $$ \frac{2}{3} \div \frac{6}{5} = \frac{2}{3} \times \frac{5}{6} $$ Multiply the fractions: $$ \frac{2}{3} \times \frac{5}{6} = \frac{2 \times 5}{3 \times 6} = \frac{10}{18} $$ Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 2: $$ \frac{10}{18} = \frac{10 \div 2}{18 \div 2} = \frac{5}{9} $$

6. Calculate the upper bound for the perimeter of the square

The length of the side is 12 cm, correct to the nearest centimetre. The upper bound for the side length is $12 + 0.5 = 12.5$ cm. The perimeter of a square is $4 \times \text{side length}$. The upper bound for the perimeter is $4 \times 12.5 = 50$ cm.

7. Work out the measurement Jo records

The upper bound for Jo's measurement is 12.35 m. Since she measures to the nearest 10 cm (which is 0.1 m), the actual measurement is $x$. The upper bound of $x$ is $x + 0.05 = 12.35$. Solving for $x$: $$ x = 12.35 - 0.05 = 12.30 $$ Since she measures to the nearest 10 cm, she records 12.3 m.