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Alex73 [517]
3 years ago
15

Determine whether the vectors u and v are parallel, orthogonal, or neither.

Mathematics
2 answers:
solong [7]3 years ago
7 0
If the vectors are parallel:
→       →
u = k · v
8 = 10 k
4 = 7  k.
If they are orthogonal:
→  →
u ·  v = 0
( 10 · 8 ) + ( 7 · 4 ) 0 80 + 28 = 108 ≠ 0
Answer: the vectors u and v are neither parallel nor orthogonal.

NISA [10]3 years ago
6 0

Answer:

The vectors are neither parallel nor orthogonal.

Step-by-step explanation:

We are given the vectors as,

u = <8,4> and v = <10,7>

So, their dot product is given by,

u · v = <8,4> · <10,7> = 8×10 + 4×7 = 80 + 28 = 108 ≠ 0

<em>As, we know, Two vectors are orthogonal if their dot product is 0.</em>

Since, u · v = 108 ≠ 0

Thus, they are not orthogonal.

<em>Also, Two vectors are parallel if angle between them is 0° or 180°.</em>

So, we will find the angle between u and v.

As, ║u║ = \sqrt{8^{2}+4^{2}} = \sqrt{64+16} = \sqrt{80} = 8.94

As, ║v║ = \sqrt{10^{2}+7^{2}} = \sqrt{100+49} = \sqrt{149} = 12.2

So, we have,

\theta = \cos^{-1} \frac{u\cdot v}{\left \| u \right \|\times \left \| v \right \|}

i.e. \theta = \cos^{-1} \frac{108}{8.94\times 12.2|}

i.e. \theta = \cos^{-1} \frac{108}{109.068}

i.e. \theta = \cos^{-1} 0.9902

i.e. \theta = 8.03

Thus, θ ≈ 8.03°, which is not equal to 0° or 180°.

Then, the vectors are not parallel.

Hence, we see that the vectors are neither parallel nor orthogonal.

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Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}

Where:

y_{o} - Initial height of the ball, measured in feet.

v_{o} - Initial speed of the ball, measured in feet per second.

g - Gravitational constant, equal to -32.174\,\frac{ft}{s^{2}}.

t - Time, measured in seconds.

Given that y_{o} = 0\,ft, v_{o} = 95\,\frac{ft}{s}, g = -32.174\,\frac{ft}{s^{2}} and y = 120\,ft, the following second-order polynomial is found:

120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}

-16.087\cdot t^{2} + 95\cdot t -120 =0

The roots of this polynomial are, respectively:

t_{1} \approx 4.075\,s and t_{2} \approx 1.831\,s.

Both roots solutions are physically reasonable, since t_{1} represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas t_{2} represents the instant when the ball the same height after reaching maximum height.

In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.

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</span>
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