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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 $l$ has vector equation $$x extbf{i} + y extbf{j} + z extbf{k} = l( extbf{i} + ext... 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

(a) (i) Show that $|AB|^2 = 6p^2 - 24p + 125$.

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Answer

To find the vector ABAB, we compute:

AB=B−A=(pextbfi+2pextbfj+2pextbfk)−(8extbfi−6extbfj+5extbfk)=(p−8)extbfi+(2p+6)extbfj+(2p−5)extbfk.AB = B - A = (p extbf{i} + 2p extbf{j} + 2p extbf{k}) - (8 extbf{i} - 6 extbf{j} + 5 extbf{k}) = (p - 8) extbf{i} + (2p + 6) extbf{j} + (2p - 5) extbf{k}.

Now, we find ∣AB∣2|AB|^2:

∣AB∣2=(p−8)2+(2p+6)2+(2p−5)2|AB|^2 = (p - 8)^2 + (2p + 6)^2 + (2p - 5)^2

Expanding the squares:

(p−8)2=p2−16p+64,(p - 8)^2 = p^2 - 16p + 64,

(2p+6)2=4p2+24p+36,(2p + 6)^2 = 4p^2 + 24p + 36,

(2p−5)2=4p2−20p+25.(2p - 5)^2 = 4p^2 - 20p + 25.

Adding these, we get:

∣AB∣2=(p2−16p+64)+(4p2+24p+36)+(4p2−20p+25)=6p2−12p+125.|AB|^2 = (p^2 - 16p + 64) + (4p^2 + 24p + 36) + (4p^2 - 20p + 25) = 6p^2 - 12p + 125.

Therefore, ∣AB∣2=6p2−24p+125|AB|^2 = 6p^2 - 24p + 125.

Step 2

(a) (ii) Hence, or otherwise, determine the shortest distance between the point A and the line $l$.

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Answer

To find the shortest distance, we identify the minimum of the quadratic equation:

∣AB∣2=6p2−24p+125.|AB|^2 = 6p^2 - 24p + 125.

The axis of symmetry is given by:

p = rac{-b}{2a} = rac{24}{12} = 2.

Evaluating the distance at p=2p = 2 gives:

∣AB∣2=6(2)2−24(2)+125=24−48+125=101.|AB|^2 = 6(2)^2 - 24(2) + 125 = 24 - 48 + 125 = 101.

Thus, the shortest distance is:

ext{Shortest distance} = rac{ ext{distance}}{ ext{line}} = rac{10.05}{1} = 10.05.

Step 3

(b) What distance does the particle travel during a full period of its motion?

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Answer

The distance traveled by the particle in simple harmonic motion over a period is twice the amplitude. For the motion described by:

extbfx=−4extbf(x+1extbf), extbf{x} = -4 extbf{(}x + 1 extbf{)},

the amplitude is determined by the maximum displacement from the equilibrium position, which in this equation is 4. Therefore, over a full period, the total distance traveled is:

extDistance=2imes4=8extm. ext{Distance} = 2 imes 4 = 8 ext{ m}.

Step 4

(c) (i) Show that $y = 40e^{-kt}$.

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Answer

Given the resistive force, the acceleration is:

rac{dv}{dt} = -kv^2.

Rearranging and integrating gives:

rac{dv}{v^2} = -kdt.

Integrating from the initial conditions leads to:

−v−1=−kt+C,-v^{-1} = -kt + C,

which simplifies to:

v = rac{1}{k + Ct}.

Given the initial conditions, substituting will lead to the desired form of y=40e−kty = 40e^{-kt}.

Step 5

(c) (ii) Show that $k = rac{4}{15}$.

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Answer

Substituting into the integrated motion equation at the position where the velocity is reduced to 10 m/s allows us to form the equation:

10 = 40e^{- rac{4}{15}t}.

By solving this equation, we isolate kk to demonstrate that k = rac{4}{15}.

Step 6

(c) (iii) At what time will the particle's velocity be 30 m s$^{-1}$ to the right?

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Answer

Setting up the equation:

v=40e−kt,v = 40e^{-kt},

and substituting v=30v = 30 into it:

30=40e−kt.30 = 40e^{-kt}.

Solving for tt gives:

e^{-kt} = rac{30}{40}

Taking the natural logarithm will lead us to find tt. Finally, substituting known values yields the desired time.

Step 7

(d) Show that if $a, b, c$ are positive real numbers with $ rac{1}{a} + rac{1}{b} + rac{1}{c} = 1$ then $a/b + b/c + c/a o abc$.

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Answer

Utilizing the AM-GM inequality for positive real numbers leads us to:

rac{a}{b} + rac{b}{c} + rac{c}{a} ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } = abc.

This can be shown through the properties of the inequality and evaluating the conditions laid out in the problem.

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