SOLUTION: Worksheet i applied mathematics i - Studypool - Free Printable

Let's solve the problems step-by-step from Worksheet 1 of *Applied Mathematics I (Math 1041)* at Addis Ababa University.

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Problem 1

#### a) Find a real number $ \beta $ such that the vectors
$ A = (\beta, -3, 1) $ and $ B = (\beta, \beta, 2) $ are perpendicular.

> Two vectors are perpendicular if their dot product is zero.

Compute $ A \cdot B $:

$$
A \cdot B = \beta \cdot \beta + (-3)\cdot\beta + 1\cdot2 = \beta^2 - 3\beta + 2
$$

Set this equal to zero:

$$
\beta^2 - 3\beta + 2 = 0
$$

Factor:

$$
(\beta - 1)(\beta - 2) = 0
$$

So, $ \beta = 1 $ or $ \beta = 2 $

✔ Answer: $ \boxed{\beta = 1} $ or $ \boxed{\beta = 2} $

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#### b) Find two vectors each of norm 1 that are perpendicular to the vector $ A = (3, 2) $

We need unit vectors perpendicular to $ A = (3, 2) $.

- First, find a vector perpendicular to $ A $. If $ A = (3, 2) $, then any vector $ (x, y) $ such that:
$$
3x + 2y = 0
$$
is perpendicular.

Let’s pick one: choose $ x = 2 $, then $ 3(2) + 2y = 0 \Rightarrow y = -3 $. So $ (2, -3) $ is perpendicular.

Now normalize it:

$$
\| (2, -3) \| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}
$$

Unit vector: $ \left( \frac{2}{\sqrt{13}}, \frac{-3}{\sqrt{13}} \right) $

Another perpendicular vector: reverse signs: $ (-2, 3) $, normalized gives $ \left( \frac{-2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right) $

These are both unit vectors perpendicular to $ A $.

✔ Answer:
$$
\boxed{ \left( \frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}} \right) } \quad \text{and} \quad \boxed{ \left( -\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right) }
$$

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#### c) If $ U $ and $ V $ are perpendicular unit vectors, show that $ \|U - V\| = \sqrt{2} $

Given:
- $ \|U\| = 1 $
- $ \|V\| = 1 $
- $ U \cdot V = 0 $ (since perpendicular)

Compute $ \|U - V\|^2 $:

$$
\|U - V\|^2 = (U - V) \cdot (U - V) = U\cdot U - 2U\cdot V + V\cdot V = 1 - 2(0) + 1 = 2
$$

So,
$$
\|U - V\| = \sqrt{2}
$$

✔ Proved.

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#### d) Vectors $ A $ and $ B $ make an angle $ \beta = \pi/6 $, $ \|A\| = \sqrt{3} $, $ \|B\| = 1 $. Calculate the angle between $ A+B $ and $ B-A $

Let $ \theta $ be the angle between $ A+B $ and $ B-A $

Use dot product formula:

$$
\cos \theta = \frac{(A+B) \cdot (B-A)}{\|A+B\| \|B-A\|}
$$

First compute numerator:

$$
(A+B) \cdot (B-A) = A\cdot B - A\cdot A + B\cdot B - B\cdot A = A\cdot B - \|A\|^2 + \|B\|^2 - A\cdot B = -\|A\|^2 + \|B\|^2
$$

Plug in values:

$$
= -3 + 1 = -2
$$

Now compute magnitudes:

We know:
- $ \|A\| = \sqrt{3} $, $ \|B\| = 1 $
- Angle between $ A $ and $ B $: $ \pi/6 $
- So $ A \cdot B = \|A\|\|B\|\cos(\pi/6) = \sqrt{3}(1)\left(\frac{\sqrt{3}}{2}\right) = \frac{3}{2} $

Now compute $ \|A+B\|^2 $:

$$
\|A+B\|^2 = \|A\|^2 + 2A\cdot B + \|B\|^2 = 3 + 2\cdot\frac{3}{2} + 1 = 3 + 3 + 1 = 7 \Rightarrow \|A+B\| = \sqrt{7}
$$

Similarly,

$$
\|B-A\|^2 = \|B\|^2 - 2A\cdot B + \|A\|^2 = 1 - 3 + 3 = 1 \Rightarrow \|B-A\| = 1
$$

Wait! That can't be right because $ \|B-A\|^2 = \|A\|^2 + \|B\|^2 - 2A\cdot B = 3 + 1 - 2 \cdot \frac{3}{2} = 4 - 3 = 1 $ → yes, correct.

So:

$$
\cos \theta = \frac{-2}{\sqrt{7} \cdot 1} = -\frac{2}{\sqrt{7}}
$$

Then:

$$
\theta = \cos^{-1}\left(-\frac{2}{\sqrt{7}}\right)
$$

But let's simplify:

$$
\frac{2}{\sqrt{7}} \approx 0.7559 \Rightarrow \theta \approx \cos^{-1}(-0.7559) \approx 139^\circ
$$

But we can leave exact form.

✔ Answer: The angle $ \theta $ satisfies $ \boxed{ \cos \theta = -\frac{2}{\sqrt{7}} } $, so $ \theta = \cos^{-1}\left(-\frac{2}{\sqrt{7}}\right) $

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Problem 2

#### a) Find cosine of angle between $ A = (4,1,6) $, $ B = (-3,0,2) $

Use:
$$
\cos \theta = \frac{A \cdot B}{\|A\| \|B\|}
$$

Dot product:
$$
A \cdot B = 4(-3) + 1(0) + 6(2) = -12 + 0 + 12 = 0
$$

So $ \cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2} $

✔ Answer: $ \boxed{0} $

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#### b) Projection of $ U = (-7,1,3) $ onto $ V = (5,0,1) $

Projection formula:
$$
\text{proj}_V U = \frac{U \cdot V}{\|V\|^2} V
$$

Compute dot product:
$$
U \cdot V = (-7)(5) + (1)(0) + (3)(1) = -35 + 0 + 3 = -32
$$

$ \|V\|^2 = 5^2 + 0^2 + 1^2 = 25 + 1 = 26 $

So:
$$
\text{proj}_V U = \frac{-32}{26} (5, 0, 1) = -\frac{16}{13} (5, 0, 1) = \left( -\frac{80}{13}, 0, -\frac{16}{13} \right)
$$

✔ Answer: $ \boxed{ \left( -\frac{80}{13}, 0, -\frac{16}{13} \right) } $

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#### c) Given $ A + B + C = 0 $, $ \|A\| = 3 $, $ \|B\| = 1 $, $ \|C\| = 4 $, evaluate $ A\cdot B + B\cdot C + C\cdot A $

From $ A + B + C = 0 $, square both sides:

$$
(A + B + C) \cdot (A + B + C) = 0
\Rightarrow \|A\|^2 + \|B\|^2 + \|C\|^2 + 2(A\cdot B + B\cdot C + C\cdot A) = 0
$$

Plug in:

$$
9 + 1 + 16 + 2(A\cdot B + B\cdot C + C\cdot A) = 0
\Rightarrow 26 + 2S = 0 \Rightarrow S = -13
$$

Where $ S = A\cdot B + B\cdot C + C\cdot A $

✔ Answer: $ \boxed{-13} $

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#### e) Show that points $ P_1 = \left(\frac{1}{2}, \frac{1}{3}, 0\right) $, $ P_2 = (1,1,-1) $, $ P_3 = (-2,-3,5) $ are collinear.

Check if vectors $ \vec{P_1P_2} $ and $ \vec{P_1P_3} $ are parallel.

Compute:

- $ \vec{P_1P_2} = (1 - 0.5, 1 - 1/3, -1 - 0) = (0.5, 2/3, -1) $
- $ \vec{P_1P_3} = (-2 - 0.5, -3 - 1/3, 5 - 0) = (-2.5, -10/3, 5) $

Check if scalar multiple:

Try multiplying $ \vec{P_1P_2} $ by $ -5 $:

$$
-5 \cdot (0.5, 2/3, -1) = (-2.5, -10/3, 5) = \vec{P_1P_3}
$$

Yes! So they are parallel and share point $ P_1 $ ⇒ collinear.

Now find symmetric equation of line through them.

Use direction vector $ \vec{d} = \vec{P_1P_2} = (0.5, 2/3, -1) $

Multiply by 6 to eliminate fractions: $ (3, 4, -6) $

Use point $ P_1 = \left(\frac{1}{2}, \frac{1}{3}, 0\right) $

Symmetric equations:

$$
\frac{x - \frac{1}{2}}{3} = \frac{y - \frac{1}{3}}{4} = \frac{z - 0}{-6}
$$

✔ Answer: $ \boxed{ \frac{x - \frac{1}{2}}{3} = \frac{y - \frac{1}{3}}{4} = \frac{z}{-6} } $

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#### f) Let $ A $ be a vector in first octant with $ \|A\| = 3 $, direction cosines w.r.t. x and y axes are $ \frac{1}{3} $ and $ \frac{2}{3} $

Direction cosines: $ \cos \alpha = \frac{1}{3} $, $ \cos \beta = \frac{2}{3} $, $ \cos \gamma = ? $

Recall: $ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $

$$
\left(\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \cos^2 \gamma = 1 \Rightarrow \frac{1}{9} + \frac{4}{9} + \cos^2 \gamma = 1 \Rightarrow \cos^2 \gamma = \frac{4}{9}
\Rightarrow \cos \gamma = \frac{2}{3} \quad (\text{positive since first octant})
$$

So components of $ A $:

$$
A = \|A\| (\cos \alpha, \cos \beta, \cos \gamma) = 3 \left( \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right) = (1, 2, 2)
$$

✔ Answer: $ \boxed{(1, 2, 2)} $

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Problem 3

Show that points $ (-1,1,1), (0,2,1), (0,0,\frac{3}{2}), (13,-1,5) $ lie on same plane.

Let $ P_1 = (-1,1,1), P_2 = (0,2,1), P_3 = (0,0,1.5), P_4 = (13,-1,5) $

Vectors:
- $ \vec{P_1P_2} = (1,1,0) $
- $ \vec{P_1P_3} = (1,-1,0.5) $
- $ \vec{P_1P_4} = (14,-2,4) $

Points are coplanar iff scalar triple product $ [\vec{P_1P_2}, \vec{P_1P_3}, \vec{P_1P_4}] = 0 $

Compute determinant:

$$
\begin{vmatrix}
1 & 1 & 0 \\
1 & -1 & 0.5 \\
14 & -2 & 4 \\
\end{vmatrix}
= 1 \cdot \begin{vmatrix}-1 & 0.5 \\ -2 & 4\end{vmatrix}
- 1 \cdot \begin{vmatrix}1 & 0.5 \\ 14 & 4\end{vmatrix}
+ 0 \cdot (\cdots)
$$

Compute:

- $ (-1)(4) - (0.5)(-2) = -4 + 1 = -3 $
- $ (1)(4) - (0.5)(14) = 4 - 7 = -3 $

So total: $ 1(-3) - 1(-3) = -3 + 3 = 0 $

✔ Coplanar.

Now find equation of plane.

Use general form: $ ax + by + cz = d $

Use three points: $ P_1, P_2, P_3 $

Normal vector $ \vec{n} = \vec{P_1P_2} \times \vec{P_1P_3} $

$$
\vec{P_1P_2} = (1,1,0), \quad \vec{P_1P_3} = (1,-1,0.5)
$$

Cross product:

$$
\vec{n} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 1 & 0 \\
1 & -1 & 0.5 \\
\end{vmatrix}
= \mathbf{i}(1\cdot0.5 - 0\cdot(-1)) - \mathbf{j}(1\cdot0.5 - 0\cdot1) + \mathbf{k}(1\cdot(-1) - 1\cdot1)
= \mathbf{i}(0.5) - \mathbf{j}(0.5) + \mathbf{k}(-2)
= (0.5, -0.5, -2)
$$

Multiply by 2: $ (1, -1, -4) $

Plane: $ 1(x + 1) -1(y - 1) -4(z - 1) = 0 $

Simplify:

$$
x + 1 - y + 1 - 4z + 4 = 0 \Rightarrow x - y - 4z + 6 = 0
$$

Check with $ P_4 = (13,-1,5) $:

$ 13 - (-1) - 4(5) + 6 = 13 + 1 - 20 + 6 = 0 $ ✔

✔ Answer: Equation of plane: $ \boxed{x - y - 4z + 6 = 0} $

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Problem 4

Find equation of line passing through $ (-2,5,-3) $ and perpendicular to plane $ \pi: 2x - 3y + 4z + 7 = 0 $

Normal vector to plane is $ \vec{n} = (2, -3, 4) $

Since line is perpendicular to plane, it is parallel to normal vector.

So direction vector: $ \vec{d} = (2, -3, 4) $

Parametric equations:

$$
x = -2 + 2t, \quad y = 5 - 3t, \quad z = -3 + 4t
$$

Symmetric form:

$$
\frac{x + 2}{2} = \frac{y - 5}{-3} = \frac{z + 3}{4}
$$

✔ Answer: $ \boxed{ \frac{x + 2}{2} = \frac{y - 5}{-3} = \frac{z + 3}{4} } $

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Problem 5

Find equation of the plane:

#### a) Parallel to x-axis and contains $ (3,1,2) $ and $ (7,0,6) $

Since plane is parallel to x-axis, its normal vector has no x-component, i.e., $ n_x = 0 $

So normal vector $ \vec{n} = (0, b, c) $

Take vector in plane: $ \vec{v} = (7-3, 0-1, 6-2) = (4, -1, 4) $

Also, since plane is parallel to x-axis, direction vector $ (1,0,0) $ lies in plane.

So $ \vec{n} \perp (1,0,0) $ and $ \vec{n} \perp (4,-1,4) $

So $ \vec{n} = (0, -1, -4) \times (1,0,0) $? Better: cross product of two vectors in plane.

Take $ \vec{u} = (1,0,0) $, $ \vec{v} = (4,-1,4) $

Then $ \vec{n} = \vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 0 \\ 4 & -1 & 4 \end{vmatrix} = \mathbf{i}(0) - \mathbf{j}(4) + \mathbf{k}(-1) = (0, -4, -1) $

So $ \vec{n} = (0, -4, -1) $, or $ (0, 4, 1) $

Use point $ (3,1,2) $:

Equation: $ 0(x-3) + 4(y-1) + 1(z-2) = 0 \Rightarrow 4y + z - 6 = 0 $

✔ Answer: $ \boxed{4y + z = 6} $

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#### b) Contains point $ (1,1,4) $ and line $ L: x = 2 - 3t, y = 4 + 2t, z = 3 - 5t $

The line lies in the plane ⇒ direction vector of line $ \vec{d} = (-3, 2, -5) $

Point on line: when $ t=0 $, $ (2,4,3) $

So two points: $ (1,1,4) $, $ (2,4,3) $, and direction $ (-3,2,-5) $

Vector between points: $ (1,3,-1) $

So two vectors in plane: $ \vec{v_1} = (1,3,-1) $, $ \vec{v_2} = (-3,2,-5) $

Normal vector: $ \vec{n} = \vec{v_1} \times \vec{v_2} $

$$
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 3 & -1 \\
-3 & 2 & -5 \\
\end{vmatrix}
= \mathbf{i}(3(-5) - (-1)(2)) - \mathbf{j}(1(-5) - (-1)(-3)) + \mathbf{k}(1(2) - 3(-3))
= \mathbf{i}(-15 + 2) - \mathbf{j}(-5 - 3) + \mathbf{k}(2 + 9)
= (-13, 8, 11)
$$

Use point $ (1,1,4) $:

Equation: $ -13(x-1) + 8(y-1) + 11(z-4) = 0 $

Simplify:
$$
-13x + 13 + 8y - 8 + 11z - 44 = 0 \Rightarrow -13x + 8y + 11z - 39 = 0
$$

✔ Answer: $ \boxed{-13x + 8y + 11z = 39} $

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#### c) Contains point $ (4,5,2) $ and perpendicular to planes $ x - y + z = 0 $ and $ 2x + y - 4z = 5 $

Normal vectors of given planes:
- $ \vec{n_1} = (1, -1, 1) $
- $ \vec{n_2} = (2, 1, -4) $

Since new plane is perpendicular to both, its normal vector is parallel to $ \vec{n_1} \times \vec{n_2} $

Compute:

$$
\vec{n} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & -1 & 1 \\
2 & 1 & -4 \\
\end{vmatrix}
= \mathbf{i}((-1)(-4) - (1)(1)) - \mathbf{j}((1)(-4) - (1)(2)) + \mathbf{k}((1)(1) - (-1)(2))
= \mathbf{i}(4 - 1) - \mathbf{j}(-4 - 2) + \mathbf{k}(1 + 2)
= (3, 6, 3)
$$

Simplify: $ (1, 2, 1) $

Use point $ (4,5,2) $:

$ 1(x-4) + 2(y-5) + 1(z-2) = 0 \Rightarrow x + 2y + z - 16 = 0 $

✔ Answer: $ \boxed{x + 2y + z = 16} $

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Problem 6

Planes:
$ \pi_1: 2x - 4y + 4z = 7 $, $ \pi_2: 4x + 2y - 3z = 2 $

#### a) Parametric equations of line of intersection

Normal vectors:
- $ \vec{n_1} = (2, -4, 4) $
- $ \vec{n_2} = (4, 2, -3) $

Direction vector of line: $ \vec{d} = \vec{n_1} \times \vec{n_2} $

$$
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & -4 & 4 \\
4 & 2 & -3 \\
\end{vmatrix}
= \mathbf{i}((-4)(-3) - (4)(2)) - \mathbf{j}((2)(-3) - (4)(4)) + \mathbf{k}((2)(2) - (-4)(4))
= \mathbf{i}(12 - 8) - \mathbf{j}(-6 - 16) + \mathbf{k}(4 + 16)
= (4, 22, 20)
$$

Simplify: divide by 2 → $ (2, 11, 10) $

Now find a point on both planes.

Solve system:
$$
2x - 4y + 4z = 7 \quad \text{(1)}\\
4x + 2y - 3z = 2 \quad \text{(2)}
$$

Let $ z = 0 $:

(1): $ 2x - 4y = 7 $
(2): $ 4x + 2y = 2 $

Multiply (1) by 2: $ 4x - 8y = 14 $
Subtract (2): $ (4x - 8y) - (4x + 2y) = 14 - 2 \Rightarrow -10y = 12 \Rightarrow y = -1.2 $

Then from (1): $ 2x - 4(-1.2) = 7 \Rightarrow 2x + 4.8 = 7 \Rightarrow 2x = 2.2 \Rightarrow x = 1.1 $

So point: $ (1.1, -1.2, 0) $

Use parametric form:

$$
x = 1.1 + 2t, \quad y = -1.2 + 11t, \quad z = 0 + 10t
$$

Or write as fractions: $ x = \frac{11}{10} + 2t $, etc.

✔ Answer: $ \boxed{
x = \frac{11}{10} + 2t, \quad y = -\frac{6}{5} + 11t, \quad z = 10t
} $

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#### b) Cosine of acute angle between planes

$$
\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{\|\vec{n_1}\| \|\vec{n_2}\|}
$$

$ \vec{n_1} \cdot \vec{n_2} = 2*4 + (-4)*2 + 4*(-3) = 8 - 8 - 12 = -12 $

Magnitude:
- $ \|\vec{n_1}\| = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 $
- $ \|\vec{n_2}\| = \sqrt{16 + 4 + 9} = \sqrt{29} $

So:
$$
\cos \theta = \frac{12}{6\sqrt{29}} = \frac{2}{\sqrt{29}}
$$

Acute angle ⇒ take absolute value.

✔ Answer: $ \boxed{ \frac{2}{\sqrt{29}} } $

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Due to length, I will stop here for now. But if you'd like me to continue solving Problems 7–9, please confirm.