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Vectors, EXT1 V1 EQ-Bank 8

In the diagram, \(OABC\) is a parallelogram where  \(\overrightarrow{O A}=\underset{\sim}{a}\)  and  \(\overrightarrow{O C}=\underset{\sim}{c}\).

Point \(X\) divides \(O C\) in ratio \(1: 3\) and point \(Y\) divides \(A C\) in the ratio \(4: 3\), as shown.
 

Prove that points \(X, Y\) and \(B\) are collinear.   (4 marks)

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\(\text{Proof (See worked solution)}\)

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\(X, Y, B \ \ \text{are collinear if:}\)

\(m_{XY}=m_{YB} \ \ \text {or} \ \ \overrightarrow{XY}=\lambda \overrightarrow{YB}\ \ …\ (1)\)

\(\overrightarrow{XY}=\overrightarrow{XC}+\overrightarrow{CY}\)

  \(\overrightarrow{XC}=\dfrac{3}{4} \overrightarrow{OC}=\dfrac{3}{4} \underset{\sim}{c}\)

  \(\overrightarrow{CY}=\dfrac{3}{7} \overrightarrow{CA}=\dfrac{3}{7}(\underset{\sim}{\alpha}-\underset{\sim}{c})\)

\(\overrightarrow{XY}=\dfrac{3}{4} \underset{\sim}{c}+\dfrac{3}{7} \underset{\sim}{a}-\dfrac{3}{7} \underset{\sim}{c}=\dfrac{9}{28} \underset{\sim}{c}+\dfrac{3}{7} \underset{\sim}{a}\)
 

\(\overrightarrow{YB}=\overrightarrow{Y A}+\overrightarrow{A B}\)

  \(\overrightarrow{YA}=\dfrac{4}{7} \overrightarrow{C A}=\dfrac{4}{7}( \underset{\sim}{a}- \underset{\sim}{c})\)

  \(\overrightarrow{AB}=\overrightarrow{OC}=\underset{\sim}{c} \quad \text{(opposite sides of parallelogram)}\)

\(\overrightarrow{Y B}=\dfrac{4}{7}(\underset{\sim}{a}-\underset{\sim}{c})+\underset{\sim}{c}=\dfrac{4}{7} \underset{\sim}{a}+\dfrac{3}{7}\underset{\sim}{c}\)
 

\(\overrightarrow{XY}\) \(=\dfrac{9}{28} \underset{\sim}{c}+\dfrac{3}{7} \underset{\sim}{a}\)
  \(=\dfrac{3}{4}\left(\dfrac{3}{7} \underset{\sim}{c}+\dfrac{4}{7} \underset{\sim}{a}\right)\)
  \(=\dfrac{3}{4} \overrightarrow{YB}\)

 
\(\therefore X, Y \ \text{and} \ B  \ \text{are collinear (see (1) above)}\)

Vectors, EXT1 V1 EQ-Bank 7

Points \(A(3,-5), B(2,3)\) and \(C(7,-1)\) form a triangle \(ABC\) in the Cartesian Plane.

  1. Find the vectors \(\overrightarrow{A C}\) and \(\overrightarrow{B C}\).    (1 mark)

  2. Find the angle  \(\angle B C A\),  giving your answer to the nearest degree.     (2 marks)

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a.     
           

\(\displaystyle \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\binom{7}{-1}-\binom{3}{-5}=\binom{4}{4}\)

\(\displaystyle \overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}=\binom{7}{-1}-\binom{2}{3}=\binom{5}{-4}\)

b.    \(\displaystyle \overrightarrow{AC} \cdot \overrightarrow{B C}=\binom{4}{4}\binom{5}{-4}=20-16=4\)

\(\abs{\overrightarrow{AC}}=\sqrt{16+16}=\sqrt{32}\)

\(\abs{\overrightarrow{BC}}=\sqrt{25+16}=\sqrt{41}\)

\(\cos \angle B C A=\dfrac{\overrightarrow{A C} \cdot \overrightarrow{B C}}{\abs{\overrightarrow{A C}}\abs{\overrightarrow{B C}}}=\dfrac{4}{\sqrt{32} \sqrt{41}}\)

\(\angle B C A=\cos ^{-1}\left(\dfrac{4}{\sqrt{32} \sqrt{41}}\right)\)

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a.      
         

\(\displaystyle \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\binom{7}{-1}-\binom{3}{-5}=\binom{4}{4}\)

\(\displaystyle \overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}=\binom{7}{-1}-\binom{2}{3}=\binom{5}{-4}\)
 

b.    \(\displaystyle \overrightarrow{AC} \cdot \overrightarrow{B C}=\binom{4}{4}\binom{5}{-4}=20-16=4\)

\(\abs{\overrightarrow{AC}}=\sqrt{16+16}=\sqrt{32}\)

\(\abs{\overrightarrow{BC}}=\sqrt{25+16}=\sqrt{41}\)

\(\cos \angle B C A=\dfrac{\overrightarrow{A C} \cdot \overrightarrow{B C}}{\abs{\overrightarrow{A C}}\abs{\overrightarrow{B C}}}=\dfrac{4}{\sqrt{32} \sqrt{41}}\)

\(\angle B C A=\cos ^{-1}\left(\dfrac{4}{\sqrt{32} \sqrt{41}}\right) = 83.65… = 84^{\circ}\ \ \text{(nearest degree)}\)

Vectors, EXT1 V1 EQ-Bank 5

A parallelogram is formed by joining the points  `A(-3,1), B(1,4), C(5,-1)` and `D(a,b)`.

Use vector methods to find `a` and `b`.   (2 marks)

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`a=1, b=-4`

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`text{Opposite sides of a parallelogram are equal and parallel.}`

`=> vec(AB)=vec(DC)`

`vec(AB)=vec(OB)-vec(OA)=((1),(4))-((-3),(1)) =((4),(3))`

`vec(DC)=vec(OC)-vec(OD)=((5),(-1))-((a),(b))=((5-a),(-1-b))`
 

`text{Equating coordinates:}`

`5-a=4\ \ =>\ \ a=1`

`-1-b=3\ \ =>\ \ b=-4`

Vectors, EXT1 V1 EQ-Bank 5 MC

In the diagram, \(OSTU\) is a parallelogram. where  \(\underset{\sim}{s}=\overrightarrow{O S}\)  and  \(\underset{\sim}{u}=\overrightarrow{O U}\).
 

Given \(M\) is the midpoint of \(\overrightarrow{OS}\), determine which expression represents vector \(\overrightarrow{MT}\).

  1. \(\underset{\sim}{s}+\dfrac{1}{2} \underset{\sim}{u}\)
  2. \(\underset{\sim}{s}-\dfrac{1}{2} \underset{\sim}{u}\)
  3. \(\dfrac{1}{2}\underset{\sim}{s}+\underset{\sim}{u}\)
  4. \(\dfrac{1}{2}\underset{\sim}{s}-\underset{\sim}{u}\)
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\(\Rightarrow C\)

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\(\overrightarrow{M T}\) \(=\overrightarrow{O T}-\overrightarrow{O M}\)  
  \(=\underset{\sim}{s}+\underset{\sim}{u}-\left(\dfrac{1}{2} \underset{\sim}{s}\right)\)  
  \(=\dfrac{1}{2}\underset{\sim}{s}+\underset{\sim}{u}\)  

 
\(\Rightarrow C\)

Vectors, EXT1 V1 EQ-Bank 26

`OABC`  is a quadrilateral.

`P`, `Q`, `R` and `S` divide each side of the quadrilateral in half as shown below.
  


 

Prove, using vectors, that  `PQRS`  is a parallelogram.  (3 marks)

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`text(See Worked Solution)`

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`text(Consider diagonal)\ \ overset(->)(OB):`

`overset(->)(OB) = overset(->)(OA) + overset(->)(AB) = overset(->)(OC)+ overset(->)(CB)`

`overset(->)(PQ) = overset(->)(PA) + overset(->)(AQ) = 1/2(overset(->)(OA) + overset(->)(AB)) = 1/2overset(->)(OB)`

`overset(->)(SR) = overset(->)(SC) + overset(->)(CR) = 1/2(overset(->)(OC) + overset(->)(CB)) = 1/2overset(->)(OB)`

`:.overset(->)(PQ) = overset(->)(SR)`

 

`text(Consider diagonal)\ \ overset(->)(AC):`

`overset(->)(AC) = overset(->)(AB) + overset(->)(BC) = overset(->)(AO) + overset(->)(OC)`

`overset(->)(QR) = overset(->)(QB) + overset(->)(BR) = 1/2(overset(->)(AB) + overset(->)(BC)) = 1/2overset(->)(AC)`

`overset(->)(PS) = overset(->)(PO) + overset(->)(OS) = 1/2(overset(->)(AO) + overset(->)(OC)) = 1/2overset(->)(AC)`

`:. overset(->)(QR) = overset(->)(PS)`

 

`text(S)text(ince)\ PQRS\ text(has equal opposite sides,)`

`PQRS\ text(is a parallelogram.)`

Vectors, EXT1 V1 EQ-Bank 24

`PQRS`  is a parallelogram, where  `overset(->)(PQ) = underset~a`  and  `vec(PS) = underset~b`
 

Prove, using vectors, that the sum of the squares of the lengths of the diagonals is equal to the sum of the squares of the lengths of the sides.  (3 marks)

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`text(See Worked Solution)`

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`overset(->)(PR) = underset~a + underset~b,\ \ overset(->)(SQ) = underset~a-underset~b`

`overset(->)(PQ) = overset(->)(SR) = underset~a`

`overset(->)(PS) = overset(->)(QR) = underset~b`

`text(Prove)\ \ |underset~a + underset~b|^2 + |underset~a-underset~b|^2 = 2|underset~a|^2 + 2|underset~b|^2`

`|underset~a + underset~b|^2 + |underset~a-underset~b|^2` `= (underset~a + underset~b) · (underset~a + underset~b) + (underset~a-underset~b)(underset~a-underset~b)`
  `= underset~a · underset~a + underset~b · underset~b + 2underset~a · underset~b + underset~a · underset~a + underset~b · underset~b-2underset~a · underset~b`
  `= |underset~a|^2 + |underset~b|^2 + |underset~a|^2 + |underset~b|^2`
  `= 2|underset~a|^2 + 2|underset~b|^2\ \ …\ text(as required)`