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topjm [15]
3 years ago
8

Find two unit vectors orthogonal to both 8, 5, 1 and −1, 1, 0 .

Mathematics
1 answer:
Elina [12.6K]3 years ago
7 0
In order to do this, you must first find the "cross product" of these vectors. To do that, we can use several methods. To simplify this first, I suggest you compute:

‹1, -1, 1› × ‹0, 1, 1›

You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s.

So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook):
http://en.wikipedia.org/wiki/Cross_produ...

In addition to these methods, sometimes I like to set up:
‹1, -1, 1› • ‹a, b, c› = 0
‹0, 1, 1› • ‹a, b, c› = 0

That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to:

a - b + c = 0
b + c = 0

This is two equations, three unknowns, so you can solve it with one free parameter:

b = -c
a = c - b = -2c

The computation, regardless of method, yields:
‹1, -1, 1› × ‹0, 1, 1› = ‹-2, -1, 1›

The above method, solving equations, works because you'd just plug in c=1 to obtain this solution. However, it is not a unit vector. There will always be two unit vectors (if you find one, then its negative will be the other of course). To find the unit vector, we need to find the magnitude of our vector:

|| ‹-2, -1, 1› || = √( (-2)² + (-1)² + (1)² ) = √( 4 + 1 + 1 ) = √6

Then we divide that vector by its magnitude to yield one solution:

‹ -2/√6 , -1/√6 , 1/√6 ›

And take the negative for the other:

‹ 2/√6 , 1/√6 , -1/√6 ›
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2 years ago
Use the following steps to prove that log b(xy)- log bx+ log by.
borishaifa [10]

Answer with Step-by-step explanation:

a.x=b^p

y=b^q

Taking both sides log

log x=plog b

Using identity:logx^y=ylogx

p=\frac{logx}{log b}=log_b x

Using identity:log_x y=\frac{log y}{log x}

log y=qlog b

q=\frac{log y}{log b}=log_b y

b.xy=b^pb^q

We know that

x^a\cdot x^b=x^{a+b}

Using identity

xy=b^{p+q}

c.log_b(xy)=log_b(b^{p+q})

log_b(xy)=(p+q)log_b b

Substitute the values then we get

log_b(xy)=(log_b x+log_b y)

By using log_b b=1

Hence, log_b(xy)=log_b x+log_b y

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3 years ago
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Solving multi step equations
hichkok12 [17]

The no-brain way to do it is to

  1. subtract one side from both sides so you have <em>(something) = 0</em>.
  2. divide by the coefficient of the variable.
  3. add the opposite of the constant.

Of course, at some point, you need to simplify the equation so you have something like

... ax + b = 0 . . . . . . where <em>a</em> and <em>b</em> are some constants that may be positive or negative

17) Subtract the right side.

... 10(x +3) -(-9x -4) -(x -5 +3) = 0

... 10x + 30 +9x +4 -x +5 -3 = 0 . . . . . eliminate parentheses

... x(10+9-1) +(30+4+5-3) = 0 . . . . . . . collect terms

... 18x +36 = 0 . . . . . . . . . . . . . . . . . . . simplified

... x + 2 = 0 . . . . . . . . . . . . . . . . . . . . . .divide by 18, the coefficient of x

... x = -2 . . . . . . . . . . . . . . . . . . . . . . . . add the opposite of the constant

19) Add the opposite of the left side.

... 0 = -9(1 +7x) +12(x -12)

... 0 = -9 -63x +12x -144 . . . . . . eliminate parenthses

... 0 = -51x -153 . . . . . . . . . . . . . simplify

... 0 = x +3 . . . . . . . . . . . . . . . . . divide by -51, the coefficient of x

... -3 = x . . . . . . . . . . . . . . . . . . . add the opposite of the constant

_____

If you examine the variable's coefficients you can make a choice of side to subtract that results in a positive coefficient of the variable.

This method puts variable and constant together until the end. The approach usually taught is to separate the variable terms and constant terms. (The number of steps required is the same either way.)

The reason this is "no brain" is that it always works and requires no judgment as to what you add or subtract from where. Applying a little judgment as described above can make it so you're mostly working with positive numbers, but the method works whether the numbers are positive or negative.

6 0
4 years ago
Let ƒ(x) = 5x − l and g(x) = x2 - 1. find (f ○ g)(−7)
attashe74 [19]

Answer:

240

Step-by-step explanation:

5(x² - 1)

5 ((-7)² - 1)

5(49 - 1)

5 (48)

4 0
3 years ago
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