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Vectors, EXT2 EQ-Bank 34

A sphere of radius  \(r=3\)  is centred at \(C(6,-3,2)\).

A line passes through \(A(3,-1,6)\) and \(B(5,-1,-5)\).

Show that this line is a tangent to the sphere.    (4 marks)

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\(\text{See Worked Solutions}\)

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\(\text{The sphere centred at} \ \ C(6,-3,2) \ \ \text {with radius} \ \ r=3\)

\((x-6)^2+(y+3)^2+(z-2)^2=3^2=9\)
 

\(\text{The line}\ A B\ \text{has direction}\)

\(\overrightarrow{O B}-\overrightarrow{O A}=\left[\begin{array}{c}5-3 \\ -1-(-1) \\ -5-6\end{array}\right]=\left[\begin{array}{c}2 \\ 0 \\ -11\end{array}\right]\)
 

\(\text{Line \(A B\) has parametric equation:}\)

\(\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\overrightarrow{O A}+\lambda \overrightarrow{A B}=\left[\begin{array}{c}3 \\ -1 \\ 6\end{array}\right]+\lambda\left[\begin{array}{c}2 \\ 0 \\ -11\end{array}\right]=\left[\begin{array}{c}3+2 \lambda \\ -1 \\ 6-11 \lambda\end{array}\right]\)
 

\(\text{Substitute into the equation for the sphere:}\)

\((3+2 \lambda-6)^2+(-1+3)^2+(6-11 \lambda-2)^2\) \(=9\)
\((2 \lambda-3)^2+4+(4-11 \lambda)^2\) \(=9\)
\(9-12 \lambda+4 \lambda^2+4+16-88 \lambda+121 \lambda^2\) \(=9\)
\(125 \lambda^2-100 \lambda+20\) \(=0\)
\(5\left(25 \lambda^2-20 \lambda+4\right)\) \(=0\)
\(5(5 \lambda-2)^2\) \(=0\)

 

\(\text{There is only one}\ \lambda\ \text{that solves this equation.}\)

\(\text{i.e. one common point on the line and the sphere.}\)

\(\therefore\ \text{The line is a tangent to the sphere.}\)

Vectors, EXT2 V1 2025 HSC 16c

Consider the point \(B\) with three-dimensional position vector \(\underset{\sim}{b}\) and the line  \(\ell: \underset{\sim}{a}+\lambda \underset{\sim}{d}\), where \(\underset{\sim}{a}\) and \(\underset{\sim}{d}\) are three-dimensional vectors, \(\abs{\underset{\sim}{d}}=1\) and \(\lambda\) is a parameter.

Let \(f(\lambda)\) be the distance between a point on the line \(\ell\) and the point \(B\).

  1. Find \(\lambda_0\), the value of \(\lambda\) that minimises \(f\), in terms of \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{d}\).   (2 marks)

  2. Let \(P\) be the point with position vector  \(\underset{\sim}{a}+\lambda_0 \underset{\sim}{d}\).
  3. Show that \(PB\) is perpendicular to the direction of the line \(\ell\).   (1 mark)

  4. Hence, or otherwise, find the shortest distance between the line \(\ell\) and the sphere of radius 1 unit, centred at the origin \(O\), in terms of \(\underset{\sim}{d}\) and \(\underset{\sim}{a}\).
  5. You may assume that if \(B\) is the point on the sphere closest to \(\ell\), then \(O B P\) is a straight line.   (3 marks)

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i.    \(\lambda_0=\underset{\sim}{d}(\underset{\sim}{b}-\underset{\sim}{a})\)

ii.   \(\text{See Worked Solutions.}\)

iii.  \(d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}\)

Show Worked Solution

i.    \(\ell=\underset{\sim}{a}+\lambda \underset{\sim}{d}, \quad\abs{\underset{\sim}{d}}=1\)

\(\text{Vector from point \(B\) to a point on \(\ell\)}:\ \underset{\sim}{a}+\lambda \underset{\sim}{d}-\underset{\sim}{b}\)

\(f(\lambda)=\text{distance between \(\ell\) and \(B\)}\)

\(f(\lambda)=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}\)

\(\text{At} \ \ \lambda_0, f(\lambda) \ \ \text{is a min}\ \Rightarrow \ f(\lambda)^2 \ \ \text {is also a min}\)

♦♦ Mean mark (i) 33%.
\(f(\lambda)^2\) \(=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}^2\)
  \(=(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})\)
  \(=(\underset{\sim}{a}-\underset{\sim}{b})\cdot (\underset{\sim}{a}-\underset{\sim}{b})+2\lambda (\underset{\sim}{a}-\underset{\sim}{b}) \cdot \underset{\sim}{d}+\lambda^2 \underset{\sim}{d} \cdot  \underset{\sim}{d}\)
  \(=\lambda^2|\underset{\sim}{d}|^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda +\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)
  \(=\lambda^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda+\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)

 

\(f(\lambda)^2 \ \ \text{is a concave up quadratic.}\)

\(f(\lambda)_{\text {min}}^2 \ \ \text{occurs at the vertex.}\)

\(\lambda_0=-\dfrac{b}{2 a}=-\dfrac{2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b})}{2}=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})\)
 

ii.    \(P \ \text{has position vector} \ \ \underset{\sim}{a}+\lambda_0 \underset{\sim}{d}\)

\(\text{Show} \ \ \overrightarrow{PB} \perp \ell:\)

♦♦♦ Mean mark (ii) 22%.

\(\overrightarrow{PB}=\underset{\sim}{b}-\underset{\sim}{p}=\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\)

\(\overrightarrow{P B} \cdot \underset{\sim}{d}\) \(=\left(\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\right) \cdot \underset{\sim}{d}\)
  \(=(\underset{\sim}{b}-\underset{\sim}{a}) \cdot \underset{\sim}{d}-\lambda_0 \underset{\sim}{d} \cdot \underset{\sim}{d}\)
  \(=\lambda_0-\lambda_0\abs{\underset{\sim}{d}}^2\)
  \(=0\)

 

\(\therefore \overrightarrow{PB}\ \text{is perpendicular to the direction of the line}\ \ell. \)
 

iii.   \(\text{Shortest distance between} \ \ell \ \text{and sphere (radius\(=1\))}\)

\(=\ \text{(shortest distance \(\ell\) to \(O\))}-1\)

♦♦♦ Mean mark (iii) 4%.

\(f\left(\lambda_0\right)=\text{shortest distance \(\ell\) to point \(B\)}\)

\(\text{Set} \ \ \underset{\sim}{b}=0 \ \Rightarrow \ f\left(\lambda_0\right)=\text{shortest distance \(\ell\) to \(0\)}\)

\(\Rightarrow \lambda_0=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})=-\underset{\sim}{d} \cdot \underset{\sim}{a}\)

\(f\left(\lambda_0\right)\) \(=\abs{\underset{\sim}{a}-\underset{\sim}{b}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}=\abs{\underset{\sim}{a}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}\)
\(f\left(\lambda_0\right)^2\) \(=\abs{\underset{\sim}{a}}^2-2( \underset{\sim}{a}\cdot \underset{\sim}{d})^2+(\underset{\sim}{d} \cdot \underset{\sim}{a})^2\abs{\underset{\sim}{d}}^2\)
  \(=\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2\)
\(f\left(\lambda_0\right)\) \(=\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}\)

 

\(\text {Shortest distance of \(\ell\) to sphere \(\left(d_{\min }\right)\):}\)

\(d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}\)

Vectors, EXT2 V1 EQ-Bank 15

  1. Find the equation of the vector line `underset~v` that passes through  `Atext{(5, 2, 3)}` and `B(7, 6, 1)`.   (1 mark)

  2. A sphere has centre `underset~c` at `text{(2, 3, 5)}` and a radius of `5sqrt2`  units.
    Find the points where the vector line `underset~v` meets the sphere.   (3 marks)

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a.    `underset~v = ((5),(2),(3)) + lambda((1),(2),(−1))`

b.    `((2),(–4),(6)), \ ((7),(6),(1))`

Show Worked Solution

a.    `overset(->)(BA) = ((7-5),(6-2),(1-3)) = ((2),(4),(−2)) = 2((1),(2),(−1))`

`underset~v = ((5),(2),(3)) + lambda((1),(2),(−1))`
 

b.    `text(General point)\ underset~v:`

`x = 5 + lambda`

`y = 2 + 2lambda`

`z = 3-lambda`
 

`text(Equation of sphere,)\ underset~c = (2, 3, 5),\ text(radius)\ 5sqrt2:`

`(x-2)^2 + (y-3)^2 + (z -5)^2` `= (5sqrt2)^2`
`(lambda + 3)^2 + (2lambda-1)^2 + (−lambda-2)^2` `= 50`
`lambda^2 + 6lambda + 9 + 4lambda^2-4lambda + 1 + lambda^2 + 4lambda + 4` `= 50`
`6lambda^2 + 6lambda + 14` `= 50`
`6lambda^2 + 6lambda-36` `= 0`
`6(lambda + 3)(lambda-2)` `= 0`
`lambda` `= –3\ text(or)\ 2`

 
`text(When)\ \ lambda = –3,`

`text(Intersection) = ((5),(2),(3))-3((1),(2),(−1)) = ((2),(–4),(6))`

`text(When)\ \ lambda = 2,`

`text(Intersection) = ((5),(2),(3)) + 2((1),(2),(−1)) = ((7),(6),(1))`

Vectors, EXT2 V1 EQ-Bank 14

A sphere is represented by the equation

`x^2-4x + y^2 + 8y + z^2-3z + 2 = 0`

  1. Determine the centre  `underset~c`  and radius of the sphere.   (2 marks)

  2. Find the vector equation of the sphere.   (1 mark)

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a.    `underset~c = ((2),(-4),({3}/{2})) \ , \ text(radius) = (9)/(2)`

b.    `| \ underset~r-((2),(-4),({3}/{2})) | = (9)/(2)`

Show Worked Solution

a.    `x^2-4x + y^2 + 8y + z^2-3z + 2 = 0`

`(x-2)^2 + (y+4)^2 + (z-{3}/{2})^2 + 2-(89)/(4) = 0`

`(x-2)^2 + (y+4)^2 + (z-{3}/{2})^2 = (81)/(4)`

`:. \ underset~c = ((2),(-4),({3}/{2})) \ , \ text(radius) = (9)/(2)`
  

b.    `text(Vector equation:)`

`| \ underset~r-((2),(-4),({3}/{2})) | = (9)/(2)`