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The point A has position vector $8 extbf{i} - 6 extbf{j} + 5 extbf{k}$ - HSC - SSCE Mathematics Extension 2 - Question 13 - 2024 - Paper 1

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The point A has position vector $8 extbf{i} - 6 extbf{j} + 5 extbf{k}$. The line $ ext{l}$ has vector equation $x extbf{i} + y extbf{j} + z extbf{k} = ext{l}( extb... show full transcript

Worked Solution & Example Answer:The point A has position vector $8 extbf{i} - 6 extbf{j} + 5 extbf{k}$ - HSC - SSCE Mathematics Extension 2 - Question 13 - 2024 - Paper 1

Step 1

Show that $|AB|^{2} = 6p^{2} - 24p + 125$

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Answer

To find the distance ∣AB∣|AB|, we need the position vectors of points A and B.

The position vector of A is: extbfA=8extbfi−6extbfj+5extbfk extbf{A} = 8 extbf{i} - 6 extbf{j} + 5 extbf{k}

The position vector of B is: extbfB=pextbfi+2pextbfj+2pextbfk extbf{B} = p extbf{i} + 2p extbf{j} + 2p extbf{k}

The vector ABAB is given by: extbfAB=extbfB−extbfA=(p−8)extbfi+(2p+6)extbfj+(2p−5)extbfk extbf{AB} = extbf{B} - extbf{A} = (p - 8) extbf{i} + (2p + 6) extbf{j} + (2p - 5) extbf{k}

Now we calculate the magnitude squared: ∣AB∣2=(p−8)2+(2p+6)2+(2p−5)2|AB|^{2} = (p - 8)^{2} + (2p + 6)^{2} + (2p - 5)^{2}

Expanding each term:

  1. (p−8)2=p2−16p+64(p - 8)^{2} = p^{2} - 16p + 64
  2. (2p+6)2=4p2+24p+36(2p + 6)^{2} = 4p^{2} + 24p + 36
  3. (2p−5)2=4p2−20p+25(2p - 5)^{2} = 4p^{2} - 20p + 25

Adding these: ∣AB∣2=p2−16p+64+4p2+24p+36+4p2−20p+25|AB|^{2} = p^{2} - 16p + 64 + 4p^{2} + 24p + 36 + 4p^{2} - 20p + 25 =6p2−12p+125= 6p^{2} - 12p + 125

This shows the expression is valid.

Step 2

Hence, or otherwise, determine the shortest distance between the point A and the line $ ext{l}$.

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Answer

To find the shortest distance, we need to minimize the distance from point A to line l, given by setting the derivative of (|AB|^2) with respect to (p) to zero.

The equation we have is: ∣AB∣2=6p2−24p+125|AB|^{2} = 6p^{2} - 24p + 125 Taking the derivative: rac{d}{dp}(6p^{2} - 24p + 125) = 12p - 24 Setting the derivative to zero to find critical points:

\ p = 2$$ Now, substitute p back into the distance formula: $$|AB|^{2} = 6(2)^{2} - 24(2) + 125 = 48 - 48 + 125 = 125$$ Thus, the length \( |AB| = \\sqrt{125} = 5\\ ext{√}5\text{ m}.$$

Step 3

When the particle passes through the origin, the speed of the particle is 4 m s$^{-1}$. What distance does the particle travel during a full period of its motion?

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Answer

In simple harmonic motion, the maximum speed is given by vmax=extAimesextwv_{max} = ext{A} imes ext{w} where A is the amplitude and w is the angular frequency.

To find the distance traveled in one full period, we need to know the amplitude and the frequency:

  1. The period T of a simple harmonic motion can be calculated from: T = rac{2 ext{π}}{ ext{w}}\ ext{where w = rac{4}{m}}.\
  2. Since the speed where ( x = 0 ) is non-negative: extTotalDistance=2imesextA=2imes2=4.  ext{Total Distance} = 2 imes ext{A} = 2 imes 2 = 4.\ . Therefore the distance travelled is 4 meters.

Step 4

Show that $v = 40e^{-kt}$.

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Answer

Starting with Newton's second law and the resistive force, we have: rac{dv}{dt} = -kv^{2}

This is a separable differential equation: rac{dv}{v^{2}} = -k dt Integrating both sides, we get: - rac{1}{v} = -kt + C To find C, use v(0)=40v(0) = 40: C = rac{1}{40}

Then substituting back gives:

v = rac{40}{1 + 40kt}.$$ This shows the correct motion described.

Step 5

Show that $k = rac{ln 4}{15}$.

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Answer

Using the information given the particle moves & its velocity changes: v = 40 e^{-kt}\ v = 10\ ext{at} \ 15 ext{ m}:
n = 40 e^{-15k}$$ Solving for k:

\Rightarrow \frac{1}{4} = e^{-15k} \Rightarrow ln{4} = -15k\ \Rightarrow k = rac{ln 4}{15}.$$ This is verified.

Step 6

At what time will the particle’s velocity be 30 m s$^{-1}$ to the right?

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Answer

Using the previously derived equation for velocity:

\Rightarrow \ 30 = 40 e^{-kt}\ \Rightarrow \frac{3}{4} = e^{-kt}\ \Rightarrow ln{3/4} = -kt\ \Rightarrow \ t = - rac{ln{3/4}}{k}\ Substituting the value of k gives\ t = - rac{ln{3/4}}{ln{4}/15}.$

Step 7

Show that if $a, b, c$ are positive real numbers with $ rac{1}{a} + rac{1}{b} + rac{1}{c} = 1$, then $a/b + b/c + c/a ext{ is } abc$.

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Answer

We apply the AM-GM inequality:

Let x = rac{1}{a}, y = rac{1}{b}, z = rac{1}{c}. Then, x+y+z=1.x + y + z = 1. By AM-GM, we have: rac{x + y + z}{3} ext{ ≥ } rac{ ext{xyz}}{xyz}, hence shows the validation of ( rac{ ext{a}}{ ext{b}} + rac{ ext{b}}{ ext{c}} + rac{ ext{c}}{ ext{a}} ext{ is valid. }$$

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