Question

1 Find the magnitude of vector
A=3i−2j+2k
Given that
A1=3i−2j+k
A1=3i−2j+k
,
A2=2i−4j−3k
A2=2i−4j−3k
,
A3=−i+2j+2k
A3=−i+2j+2k
, find the magnitudes of
2A1−3A2−5A3

3 Given that
A1=2i−j+k
A1=2i−j+k
,
A2=i+3j−2k
A2=i+3j−2k
,
A3=3i+2j+5k
A3=3i+2j+5k
and
A4=3i+2j+5k
A4=3i+2j+5k

,Find scalars a, b, c such that
A4=aA1+bA2+cA3

4 Let a and b be vectors, then
a×b=absinθ
a×b=absin⁡θ
is the €¦ €¦ €¦product
5 A car travels 3km due north, then 5km northeast. Determine the resultant displacement

Answer 1

Answer on Question #63503 – Math – Analytic Geometry

Question

1. Given:

$$A = 3i - 2j + 2k$$

Find: magnitude of $A$ (that is, $|A|$).

Solution

$$|A| = \sqrt{3^2 + (-2)^2 + 2^2} = \sqrt{9 + 4 + 4} = \sqrt{17}$$

Answer: $\sqrt{17}$

Question

2. Given:

$$A_1 = 3i - 2j + k$$ $$A_2 = 2i - 4j - 3k$$ $$A_3 = -i + 2j + 2k$$

Find: magnitude of $B = 2A_1 - 3A_2 - 5A_3$

Solution

$$B = 2A_1 - 3A_2 - 5A_3 = 2(3i - 2j + k) - 3(2i - 4j - 3k) - 5(-i + 2j + 2k) = 6i - 4j + 2k - 6i + 12j + 9k + 5i - 10j - 10k = 5i - 2j + k$$ $$|B| = \sqrt{5^2 + (-2)^2 + 1^2} = \sqrt{25 + 4 + 1} = \sqrt{30}$$

Answer: $\sqrt{30}$

Question

3. Given:

$$A_1 = 2i - j + k$$ $$A_2 = i + 3j - 2k$$ $$A_3 = 3i + 2j + 5k$$ $$A_4 = 3i + 2j + 5k$$ $$A_4 = aA_1 + bA_2 + cA_3$$

Find: $a, b, c$

Solution

$$\begin{array}{l} 3i + 2j + 5k = 2ai - aj + ak + bi + 3bj - 2bk + 3ci + 2cj + 5ck \\ 3i + 2j + 5k = (2a + b + 3c)i + (-a + 3b + 2c)j + (a - 2b + 5c)k \end{array}$$

so we obtain the system of linear equations:

$$\left\{ \begin{array}{l} 2a + b + 3c = 3 \\ -a + 3b + 2c = 2 \\ a - 2b + 5c = 5 \end{array} \right. \quad \left\{ \begin{array}{l} 1 \\ 2 \\ 3 \end{array} \right.$$ $$\begin{array}{l} \{2\} + \{3\} \Rightarrow \\ b + 7c = 7 \Rightarrow b = 7(1 - c) \\ \{2\} - 3 \cdot \{1\} \Rightarrow \\ -7a - 7c = -7 \Rightarrow a + c = 1 \Rightarrow a = 1 - c \\ \{3\} \Rightarrow \\ (1 - c) - 2 \cdot 7(1 - c) + 5c = 5 \Rightarrow \\ 1 - c - 14 + 14c + 5c = 5 \Rightarrow 18c = 18 \Rightarrow \\ c = 1 \Rightarrow a = 1 - c = 1 - 1 = 0 \Rightarrow b = 7(1 - c) = 7(1 - 1) = 0 \end{array}$$

Answer: $a = 0, \quad b = 0, \quad c = 1$.

Question

4) Given:

$$a \times b = ab \sin \theta$$

Solution

It is the formula for the magnitude of the cross product.

Answer: $|a \times b| = ab \sin \theta$

Question

5)

A car travels 3km due north, then 5km northeast. Determine the resultant displacement

Given:

$$a = 3, \quad b = 5$$

Find: $d$ resultant displacement

Solution

northeast $\Rightarrow$ angle between $a$ and $b$ is $\beta = 135{}^{\circ}$

Using the law of cosines we obtain

$$\begin{array}{l} d^2 = a^2 + b^2 - 2ab \cos \beta = 9 + 25 - 30 \cos 135{}^{\circ} = 34 + 30 \cdot \sin 45{}^{\circ} = \\ = 34 + 15\sqrt{2} \end{array}$$ $$d = \sqrt{34 + 15\sqrt{2}}$$

Answer: $\sqrt{34 + 15\sqrt{2}}$

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