Further vectors – further maths test answers to download as PDF

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1. Answer to Question 1 (5 marks)

   Method: the scalar product is \(\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta\). Two non-zero vectors are perpendicular exactly when their scalar product is 0. the vector product \(\mathbf a\times\mathbf b\) is perpendicular to both vectors and has magnitude \(|\mathbf a||\mathbf b|\sin\theta\), the area of the parallelogram with sides \(\mathbf a\) and \(\mathbf b\). Compute it as a \(3\times3\) determinant. Mark scheme as shown in brackets.

   1. \(\mathbf a\cdot\mathbf b=3+0-2=1\) [1]; \(\cos\theta=\dfrac1{\sqrt6\sqrt{13}}\) [1]; \(\theta=\) 83.5\(^\circ\) [1]
   2. \(\mathbf a\times\mathbf b=(4,-5,-6)\) [1]; area \(=\tfrac12\sqrt{16+25+36}=\tfrac12\sqrt{77}=\) 4.39 [1]

2. Answer to Question 2 (5 marks)

   Method: substitute the general point of the line into the equation of the plane to find the parameter. the angle between a line with direction \(\mathbf d\) and a plane with normal \(\mathbf n\) satisfies \(\sin\varphi=\dfrac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}\) (the complement of the angle with the normal). The angle between two planes is the angle between their normals. Mark scheme as shown in brackets.

   1. \(3(2+\lambda)+(-1+2\lambda)-(-2\lambda)=7\) [1] gives \(5+7\lambda=7\), \(\lambda=\tfrac27\) [1]. Point \(\left(\tfrac{16}7,-\tfrac37,-\tfrac47\right)\) [1]
   2. \(\sin\varphi=\dfrac{|3+2+2|}{3\sqrt{11}}\) [1] \(=0.7035\), \(\varphi=\) 44.7\(^\circ\) [1]

   Check: \(3\times\tfrac{16}7-\tfrac37+\tfrac47=\tfrac{49}7=7\), so the point lies in the plane.

3. Answer to Question 3 (4 marks)

   Method: a plane with normal \(\mathbf n\) through the point \(A\) has equation \(\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n\), that is \(n_1x+n_2y+n_3z=d\). A normal to a plane containing two directions is their vector product. Mark scheme as shown in brackets.

   1. \(\overrightarrow{AB}=(1,-1,2)\), \(\overrightarrow{AC}=(-1,1,1)\) [1]; cross product \(=(-3,-3,0)\) [1]; plane \(x+y=2\) (using \(A\)) [1]
   2. \(\dfrac{|3+3-2|}{\sqrt2}=\dfrac4{\sqrt2}=\) \(2\sqrt2\) [1]

   Check: \(B\): \(2+0=2\) and \(C\): \(0+2=2\). The plane is vertical (it contains the \(z\) direction).

4. Answer to Question 4 (3 marks)

   Method: the distance from \(P\) to the line through \(A\) with direction \(\mathbf d\) is \(\dfrac{|\overrightarrow{PA}\times\mathbf d|}{|\mathbf d|}\), the height of the parallelogram with sides \(\overrightarrow{PA}\) and \(\mathbf d\). Mark scheme as shown in brackets.

   1. \(\overrightarrow{PA}=(-1,-3,-2)\) [1]; \(\overrightarrow{PA}\times\mathbf d=(-4,0,2)\) [1]; distance \(=\dfrac{\sqrt{20}}{\sqrt6}=\) 1.83 [1]

   Answer: the shortest distance is \(\sqrt{\tfrac{20}6}=\sqrt{\tfrac{10}3}\approx1.83\).

   Check with the foot of the perpendicular: \(\lambda=\dfrac{(1,3,2)\cdot(1,1,2)}6=\dfrac43\), giving the foot \(\left(\tfrac43,\tfrac73,\tfrac83\right)\) and the same distance \(\sqrt{\tfrac{10}3}\).

5. Answer to Question 5 (3 marks)

   Method: the angle between a line with direction \(\mathbf d\) and a plane with normal \(\mathbf n\) satisfies \(\sin\varphi=\dfrac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}\) (the complement of the angle with the normal). The angle between two planes is the angle between their normals. Mark scheme as shown in brackets.

   1. Normals \((3,0,4)\) and \((0,0,1)\): \(\cos\theta=\dfrac45\), \(\theta=\) 36.9\(^\circ\) [1]
   2. \(\sin\varphi=\dfrac{|-3-8|}{\sqrt5\times5}=\dfrac{11}{5\sqrt5}=0.984\) [1], \(\varphi=\) 79.7\(^\circ\) [1]

   So the sunlight strikes the panel at almost \(80^\circ\), close to the ideal \(90^\circ\).

Total : 20 marks