Question

2 Given that A1=3i−2j+k,A2=2i−4j−3k,A3=−i+2j+2k, find the magnitudes of 2A1−3A2−5A3
3 Given that A1=2i−j+k,A2=i+3j−2k,A3=3i+2j+5kand A4=3i+2j+5k,Find scalars a, b, c suchthat A4=aA1+bA2+cA3
5 A car travels 3km due north, then 5km northeast. Determine the resultant displacement
6 If A1=3i−j−4k, A2=−2i+4j−3k,A3=i+2j−k, find |3A1−2A3+4A3|

Answer 1

Answer on Question #59201 – Math – Analytic Geometry

Question

2. Given that $A_{1} = 3i - 2j + k, A_{2} = 2i - 4j - 3k, A_{3} = -i + 2j + 2k$, find the magnitude of $2A_{1} - 3A_{2} - 5A_{3}$.

Solution

$$\begin{array}{l} 2A_{1} - 3A_{2} - 5A_{3} = 2(3i - 2j + k) - 3(2i - 4j - 3k) - 5(-i + 2j + 2k) = 6i - 4j + 2k - \\ -6i + 12j + 9k + 5i - 10j - 10k = 5i - 2j + k. \end{array}$$

The magnitude of $2A_{1} - 3A_{2} - 5A_{3}$ is equal to $\sqrt{5^{2} + (-2)^{2} + 1^{2}} = \sqrt{30}$.

**Answer:** $\sqrt{30}$.

Question

3. Given that $A_{1} = 2i - j + k, A_{2} = i + 3j - 2k, A_{3} = 3i + 2j + 5k$, and $A_{4} = 3i + 2j + 5k$, find scalars $a, b, c$ such that $A_{4} = aA_{1} + bA_{2} + cA_{3}$.

Solution

First of all, we shall check that $A_{1}, A_{2}$ and $A_{3}$ are linearly independent. We rewrite them in the coordinate form: $A_{1} = (2, -1, 1), A_{2} = (1, 3, -2), A_{3} = (3, 2, 5)$. Then we calculate the determinant which consists of their coordinates:

$$\Delta = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 3 & -2 \\ 3 & 2 & 5 \end{vmatrix} = 30 + 6 + 2 - 9 + 8 + 5 = 42 \neq 0. \text{ Since } \Delta \neq 0 \text{ the vector } A_{4} \text{ can be represented as a linear combination of the vectors } A_{1}, A_{2} \text{ and } A_{3} \text{ in one way only. Since } A_{4} = A_{3}$$

we conclude that $A_{4} = 0 \cdot A_{1} + 0 \cdot A_{2} + 1 \cdot A_{3} \Leftrightarrow \begin{cases} a = 0 \\ b = 0. \\ c = 1 \end{cases}$

**Answer:** $a = 0, b = 0, c = 1$.

Question

5. A car travels $3\,\mathrm{km}$ due north, then $5\,\mathrm{km}$ northeast. Determine the resultant displacement.

Solution

The movement of a car is shown in the figure:

where $AB = 3, BC = 5, \angle ABC = 135{}^{\circ}$. The resultant displacement is $AC$.

Now we use the Cosine Rule:

$$\begin{array}{l} AC = \sqrt{AB^{2} + BC^{2} - 2 \cdot AB \cdot BC \cdot \cos \angle ABC} = \sqrt{9 + 25 - 30 \cdot \left(-\frac{\sqrt{2}}{2}\right)} = \\ = \sqrt{34 + 15\sqrt{2}} \approx 7.43 \text{ km}. \end{array}$$

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

Question

6. If $A_1 = 3i - j - 4k$, $A_2 = -2i + 4j - 3k$, $A_3 = i + 2j - k$, find $|3A_1 - 2A_2 + 4A_3|$.

Solution

$3A_{1} - 2A_{2} + 4A_{3} = 3(3i - j - 4k) - 2(-2i + 4j - 3k) + 4(i + 2j - k) = 9i - 3j - 12k + 4i - 8j + 6k + 4i + 8j - 4k = 17i - 3j - 10k.$

Then $|3A_1 - 2A_2 + 4A_3| = \sqrt{17^2 + (-3)^2 + (-10)^2} = \sqrt{398}$.

Answer: $\sqrt{398}$.

www.AssignmentExpert.com