Question

What type of vector is w+(-w)=0
null vector
scalar vector
vector
magnitude
2 A scalar quantity has ---- only.
direction
magnitude
force
vector
3 Find the angle between U =4i - 2j + 4k and V = 3i - 6j - 2k .

680

670

580

690
4 A ..... is such that it can slide along its line of action.
vector
null vector
motion
line vector
5 If U = I + 3j - 2k and V = 4i - 2j - 4k are vectors, find 3U + V
10
14
4
11

Answer 1

Answer on Question # 52238 – Math – Vector Calculus

Task:

1. What type of vector is $w + (-w) = 0$

a. null vector

b. scalar vector

c. vector

d. magnitude

**Answer:** a. null vector

2. A scalar quantity has --- only.

a. direction

b. magnitude

c. force

d. vector

**Answer:** b. magnitude

3. Find the angle between $U = 4i - 2j + 4k$ and $V = 3i - 6j - 2k$.

a. $68{}^{\circ}$

b. $67{}^{\circ}$

c. $58{}^{\circ}$

d. $69{}^{\circ}$

**Answer:** b. $67{}^{\circ}$

$$\cos \alpha = \frac{U \cdot V}{|U| \cdot |V|} = \frac{4 \cdot 3 + (-2) \cdot (-6) + 4 \cdot (-2)}{\sqrt{4^2 + (-2)^2 + 4^2} \sqrt{3^2 + (-6)^2 + (-2)^2}} = \frac{16}{42} = \frac{8}{21} \Rightarrow \alpha = \arccos \frac{8}{21} \approx 67{}^{\circ}$$

4. A ... is such that it can slide along its line of action.

a. vector

b. null vector

c. motion

d. line vector

**Answer:** a. vector

5. If $U = I + 3j - 2k$ and $V = 4i - 2j - 4k$ are vectors, find $|3U + V|$.

a. 10

b. 14

c. 4

d. 11

Answer: b.14

Method 1

$$3U + V = 3(i + 3j - 2k) + 4i - 2j - 4k = 3i + 9j - 6k + 4i - 2j - 4k = 7i + 7j - 10k$$ $$|3U + V| = \sqrt{7^2 + 7^2 + (-10)^2} = \sqrt{198} \approx 14.07$$

Method 2

$$\begin{aligned} (3U + V) \cdot (3U + V) &= (3U + V)^2 = 9U^2 + 2 \cdot 3U \cdot V + V^2 = \\ &= 9(1 \cdot 1 + 3 \cdot 3 + (-2) \cdot (-2)) + 6(1 \cdot 4 + 3 \cdot (-2) + (-2) \cdot (-4)) + \\ &\quad + (4 \cdot 4 + (-2) \cdot (-2) + (-4) \cdot (-4)) = 9(1 + 9 + 4) + 6(4 - 6 + 8) + (16 + 4 + 16) = \\ &= 9 \cdot 14 + 6 \cdot 6 + 36 = 198 \end{aligned}$$ $$|3U + V| = \sqrt{(3U + V)^2} = \sqrt{198} \approx 14.07$$

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