Question 12
A model aircraft flies from a point $O$ towards a survey point $P$, where $\overrightarrow{OP}=600\hat{\boldsymbol{\imath}}+800\hat{\boldsymbol{\jmath}}+100\hat{\boldsymbol{k}}$ (displacements in metres).
In still air, the velocity ($\text{m s}^{-1}$) of the aircraft would be $\boldsymbol{v}_{\mathrm{a}}=6\hat{\boldsymbol{\imath}}+8\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}}$. The velocity ($\text{m s}^{-1}$) of the wind is $\boldsymbol{v}_{\mathrm{w}}=1.5\hat{\boldsymbol{\imath}}-2\hat{\boldsymbol{\jmath}}$, so the velocity of the aircraft is $\boldsymbol{v}=\boldsymbol{v}_{\mathrm{a}}+\boldsymbol{v}_{\mathrm{w}}$.
Determine $\boldsymbol{v}$.
[1 mark] b)Use your result from Question 12a) to determine the scalar projection of $\boldsymbol{v}$ on $\overrightarrow{OP}$.
[2 marks] c)Determine the magnitude of the component of $\boldsymbol{v}$ that is perpendicular to $\overrightarrow{OP}$.
[2 marks]The wind pushes the aircraft off the line OP
Watch the wind add to the still-air velocity. Then watch the new velocity split into a part along the line OP and a part at right angles to it.
All the vectors here lie in one plane, so it is drawn face on and to scale. $\boldsymbol{v}$ is the hypotenuse of a right-angled triangle. Its legs are the parts of $\boldsymbol{v}$ along and across the line OP.
Find the foot of the perpendicular
Slide the point F along the line OP. Find the place where FV is perpendicular to OP. Then OF is the scalar projection and FV is the perpendicular component. Switch routes to see the working for each method in part c).
The angle at F is more than 90°, so FV is not perpendicular to OP yet. Move F further along the line.
The angle at F is less than 90°, so F has gone past the foot of the perpendicular. Move F back towards O.
FV is perpendicular to OP. OF is the scalar projection, $\frac{94}{\sqrt{101}}\approx9.35$, and $\boldsymbol{v}$ is the hypotenuse. So the perpendicular component is $\sqrt{93.25-\frac{8836}{101}}\approx2.40\text{ m s}^{-1}$.
FV is perpendicular to OP. The vector from O to F is the vector projection, and $\boldsymbol{v}_{\perp}=\boldsymbol{v}-\text{proj}$ is the vector from F to the tip of $\boldsymbol{v}$. Its magnitude is about $2.40\text{ m s}^{-1}$, the same as Method 1.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Keep unrounded values in your calculator until the last line.
Add $\boldsymbol{v}_{\mathrm{a}}$ and $\boldsymbol{v}_{\mathrm{w}}$ by matching components. The wind has no $\hat{\boldsymbol{k}}$ term, so its $\hat{\boldsymbol{k}}$ component is $0$.
$$\begin{aligned}\boldsymbol{v}&=(6\hat{\boldsymbol{\imath}}+8\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}})+(1.5\hat{\boldsymbol{\imath}}-2\hat{\boldsymbol{\jmath}})\\&=7.5\hat{\boldsymbol{\imath}}+6\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}}\text{ m s}^{-1}\end{aligned}$$
Marker: correctly determines $\boldsymbol{v}$
The scalar projection on $\overrightarrow{OP}$ is $\boldsymbol{v}\cdot\overrightarrow{OP}\div\lvert\overrightarrow{OP}\rvert$, so find both parts first.
$$\begin{aligned}\boldsymbol{v}\cdot\overrightarrow{OP}&=7.5(600)+6(800)+1(100)\\&=9400\\[4pt]\lvert\overrightarrow{OP}\rvert&=\sqrt{600^2+800^2+100^2}\approx1004.99\end{aligned}$$
Enter √(600²+800²+100²). The output is 1004.99.
Marker: correctly determines the scalar product and the magnitude of $\overrightarrow{OP}$
Divide the scalar product by the magnitude of $\overrightarrow{OP}$.
Enter (7.5×600 + 6×800 + 1×100) ÷ √(600² + 800² + 100²). Keep the full output, 9.35335, for c).
Marker: correctly determines the scalar projection
$\boldsymbol{v}$ is the hypotenuse of a right-angled triangle with legs the scalar projection and the perpendicular component $\boldsymbol{v}_{\perp}$. Use Pythagoras.
$$\begin{aligned}\lvert\boldsymbol{v}\rvert^2&=7.5^2+6^2+1^2=93.25\\\lvert\boldsymbol{v}_{\perp}\rvert&=\sqrt{93.25-9.35335^2}\end{aligned}$$
Marker: correctly uses the magnitude of the velocity and the scalar projection
Evaluate, using the unrounded scalar projection.
Enter √(93.25 − (9400 ÷ √(1010000))²). The output is 2.40101.
Using the scalar projection rounded to 9.35 gives 2.41; this is accepted. Follow-through applies in c) from an incorrect result in b).
Marker: correctly determines the magnitude of the perpendicular component
Add $\boldsymbol{v}_{\mathrm{a}}$ and $\boldsymbol{v}_{\mathrm{w}}$ by matching components. The wind has no $\hat{\boldsymbol{k}}$ term, so its $\hat{\boldsymbol{k}}$ component is $0$.
$$\begin{aligned}\boldsymbol{v}&=(6\hat{\boldsymbol{\imath}}+8\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}})+(1.5\hat{\boldsymbol{\imath}}-2\hat{\boldsymbol{\jmath}})\\&=7.5\hat{\boldsymbol{\imath}}+6\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}}\text{ m s}^{-1}\end{aligned}$$
Marker: correctly determines $\boldsymbol{v}$
The scalar projection on $\overrightarrow{OP}$ is $\boldsymbol{v}\cdot\overrightarrow{OP}\div\lvert\overrightarrow{OP}\rvert$, so find both parts first.
$$\begin{aligned}\boldsymbol{v}\cdot\overrightarrow{OP}&=7.5(600)+6(800)+1(100)\\&=9400\\[4pt]\lvert\overrightarrow{OP}\rvert&=\sqrt{600^2+800^2+100^2}\approx1004.99\end{aligned}$$
Enter √(600²+800²+100²). The output is 1004.99.
Marker: correctly determines the scalar product and the magnitude of $\overrightarrow{OP}$
Divide the scalar product by the magnitude of $\overrightarrow{OP}$.
Enter (7.5×600 + 6×800 + 1×100) ÷ √(600² + 800² + 100²). Keep the full output, 9.35335, for c).
Marker: correctly determines the scalar projection
Multiply $\overrightarrow{OP}$ by $\tfrac{9400}{1\,010\,000}$ for the vector projection, then subtract it from $\boldsymbol{v}$.
$$\begin{aligned}\text{proj}&\approx5.584\hat{\boldsymbol{\imath}}+7.446\hat{\boldsymbol{\jmath}}+0.931\hat{\boldsymbol{k}}\\\boldsymbol{v}_{\perp}=\boldsymbol{v}-\text{proj}&\approx1.916\hat{\boldsymbol{\imath}}-1.446\hat{\boldsymbol{\jmath}}+0.069\hat{\boldsymbol{k}}\end{aligned}$$
Store 9400÷1010000 as A. Then A×600, A×800, A×100 give 5.58416, 7.44554, 0.930693.
Marker: correctly uses the magnitude of the velocity and the scalar projection
Take the magnitude of the perpendicular component.
Enter √(1.91584² + 1.44554² + 0.0693069²). The output is 2.40101.
Follow-through applies in c) from an incorrect result in b).
Marker: correctly determines the magnitude of the perpendicular component
Each phrase of the question gave you something
If I had to guess where you might lose marks…
- Using $\boldsymbol{v}_{\mathrm{a}}$ instead of $\boldsymbol{v}$ in b). You get $\boldsymbol{v}_{\mathrm{a}}\cdot\overrightarrow{OP}=10\,100$, giving $\dfrac{10\,100}{1004.99}=\sqrt{101}\approx10.05$. This is just the still-air speed, because $\overrightarrow{OP}=100\,\boldsymbol{v}_{\mathrm{a}}$.
- Subtracting the wind in a). $\boldsymbol{v}_{\mathrm{a}}-\boldsymbol{v}_{\mathrm{w}}=4.5\hat{\boldsymbol{\imath}}+10\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}}$ gives a dot product of 10 800 and a scalar projection of about 10.75.
- Dividing by the wrong magnitude in b). Dividing by $\lvert\boldsymbol{v}\rvert$ gives $\dfrac{9400}{\sqrt{93.25}}\approx973.43$, the projection of $\overrightarrow{OP}$ on $\boldsymbol{v}$. Dividing by $\lvert\overrightarrow{OP}\rvert^2$ gives $\dfrac{9400}{1\,010\,000}\approx0.0093$, the multiplier for the vector projection, not a speed.
- Mishandling Pythagoras in c). $\lvert\boldsymbol{v}\rvert-9.35335\approx0.30$ subtracts lengths, not squares. $\sqrt{93.25-9.35335}\approx9.16$ forgets to square the projection. $\sqrt{93.25+87.4851}\approx13.44$ adds instead of subtracting.
- Early rounding in c). Using 9.35 gives 2.41 rather than 2.40. It is accepted, but the habit costs marks elsewhere.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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