Vectors in geometry questions with answers

Questions

1. Question 1One vector as a multiple of another

   $\overrightarrow{PQ} = 2\mathbf{a} + 5\mathbf{b}$ and $\overrightarrow{RS} = 4\mathbf{a} + 10\mathbf{b}$. Find the value of $k$ such that $\overrightarrow{RS} = k\overrightarrow{PQ}$.

   Show the worked solution

   1. Step 1: Compare the coefficients of a$k \times 2 = 4$$k = 2$
   2. Step 2: Check with the coefficients of b$2 \times 5 = 10$
2. Question 2One vector as a multiple of another

   $\overrightarrow{PQ} = 4\mathbf{a} + \mathbf{b}$ and $\overrightarrow{RS} = 2\mathbf{a} + \dfrac{1}{2}\mathbf{b}$. Find the value of $k$ such that $\overrightarrow{RS} = k\overrightarrow{PQ}$.

   Show the worked solution

   1. Step 1: Compare the coefficients of a$k \times 4 = 2$$k = \dfrac{1}{2}$
   2. Step 2: Check with the coefficients of b$\dfrac{1}{2} \times 1 = \dfrac{1}{2}$
3. Question 3Parallel or not?

   $\overrightarrow{PQ} = \mathbf{a} - 3\mathbf{b}$ and $\overrightarrow{RS} = 3\mathbf{a} - 9\mathbf{b}$. State whether $PQ$ and $RS$ are parallel.

   Show the worked solution

   1. Step 1: Compare the coefficients of a$\dfrac{3}{1} = 3$
   2. Step 2: Compare the coefficients of b$\dfrac{-9}{-3} = 3$
   3. Step 3: The same multiple: RS = k PQParallel? Yes.
4. Question 4Parallel or not?

   $\overrightarrow{PQ} = 2\mathbf{a} - \mathbf{b}$ and $\overrightarrow{RS} = -\mathbf{a} + 2\mathbf{b}$. State whether $PQ$ and $RS$ are parallel.

   Show the worked solution

   1. Step 1: Compare the coefficients of a$\dfrac{-1}{2} = -\dfrac{1}{2}$
   2. Step 2: Compare the coefficients of b$\dfrac{2}{-1} = -2$
   3. Step 3: Different multiples: RS is not a multiple of PQParallel? No.
5. Question 5Collinear points

   $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$ and $\overrightarrow{OC} = -\dfrac{1}{2}\mathbf{a} + \dfrac{3}{2}\mathbf{b}$. Find the value of $k$ such that $\overrightarrow{AC} = k\overrightarrow{AB}$.

   Show the worked solution

   1. Step 1: AB = OB − OA$\overrightarrow{AB} = -\mathbf{a} + \mathbf{b}$
   2. Step 2: AC = OC − OA$\overrightarrow{AC} = -\dfrac{1}{2}\mathbf{a} + \dfrac{3}{2}\mathbf{b} - \mathbf{a}$$\overrightarrow{AC} = -\dfrac{3}{2}\mathbf{a} + \dfrac{3}{2}\mathbf{b}$
   3. Step 3: Take out the common factor: AC is k times AB$\overrightarrow{AC} = \dfrac{3}{2}(-\mathbf{a} + \mathbf{b})$$k = \dfrac{3}{2}$
6. Question 6Collinear points

   $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$ and $\overrightarrow{OC} = 2\mathbf{a} - \mathbf{b}$. Find the value of $k$ such that $\overrightarrow{AC} = k\overrightarrow{AB}$.

   Show the worked solution

   1. Step 1: AB = OB − OA$\overrightarrow{AB} = -\mathbf{a} + \mathbf{b}$
   2. Step 2: AC = OC − OA$\overrightarrow{AC} = 2\mathbf{a} - \mathbf{b} - \mathbf{a}$$\overrightarrow{AC} = \mathbf{a} - \mathbf{b}$
   3. Step 3: Take out the common factor: AC is k times AB$\overrightarrow{AC} = -1(-\mathbf{a} + \mathbf{b})$$k = -1$