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3 years ago

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

Mathematics

2 answers:

bezimeni [28]3 years ago

6 0

The given vectors are and

Two vectors are said to be orthogonal if they are perpendicular to each other. i.e. the dot product of the two vectors is zero.

Let us consider the dot product of the vectors

$. ==$ which is not equal to zero. Hence, the given vectors are not orthogonal.

Two vectors are said to be parallel if one vector is a scalar multiple of the other vector.

Here we can observe that the from the given vectors, one vector can not be expressed as the scalar multiple of the other vector.

Hence, the given vectors are neither parallel nor orthogonal.

Gnom [1K]3 years ago

3 0

Answer:

Neither

Step-by-step explanation:

Given: The given vectors are $u=$ and $v=$ .

To find: To find whether the vectors u and v are parallel, orthogonal, or neither

Solution: Two vectors are said to be orthogonal if they are perpendicular to each other that is the dot product of the two vectors is zero.

Let us consider the dot product of the vectors

$u{\cdot}v={\cdot}$

$u{\cdot}v=$

which is not equal to zero.

Hence, the given vectors are not orthogonal.

Two vectors are said to be parallel if one vector is a scalar multiple of the other vector.

In the given vectors, it can be observed that , one vector can not be expressed as the scalar multiple of the other vector.

Hence, the given vectors are neither parallel nor orthogonal.

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I don’t really get this one. Thx and have a great day!!

gayaneshka [121]

Answer:

The correct option is C). (9,4)

The coordinates of a point N is (9,4)

Step-by-step explanation:

Theory: If point P(x,y) lies on line segment AB and AP: PB=m:n, then we say P divides line AB internally in ratio of m:n and Point is given by

P= $(\frac{mX2+nX1}{m+n} , \frac{mY2+nY1}{m+n})$

Given that point, M is lying somewhere between point L and point N.

The coordinates of a point L is (-6,14)

The coordinates of a point M is (-3,12)

Also, LM: MN = 1:4

We can write as,

Let,

Point L(-6,14)=(X1, Y1)

Point M(-3,12)=(x,y)

Point N is (X2, Y2)

m=1 and n=4

M(-3,12)= $(\frac{mX2+nX1}{m+n} , \frac{mY2+nY1}{m+n})$

M(-3,12)= $(\frac{1(X2)+4(-6)}{1+4} , \frac{(Y2)+4(14)}{1+4})$

M(-3,12)= $(\frac{(X2)-24}{5} , \frac{1(Y2)+56}{5})$

$(-3)=\frac{(X2)-24}{5}$

(-15)=X2-24

X2=9

$(12)=\frac{(Y2)+56}{5}$

(60)=Y2+56

Y2=4

Thus,

The coordinates of a point N is (9,4)

Result: The correct option is C). (9,4)

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4 years ago

D Evaluate arcsin (6)] at x = 4. dx

sineoko [7]

Answer:

$\frac{1}{2\sqrt{5} }$

Step-by-step explanation:

Let, $\text{sin}^{-1}(\frac{x}{6})$ = y

sin(y) = $\frac{x}{6}$

$\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})$

$\frac{d}{dx}\text{sin(y)}=\frac{1}{6}$

$\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx}$ ---------(1)

$\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}$

cos(y) = $\sqrt{1-\text{sin}^{2}(y) }$

= $\sqrt{1-(\frac{x}{6})^2}$

= $\sqrt{1-(\frac{x^2}{36})}$

Therefore, from equation (1),

$\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

Or $\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

At x = 4,

$\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}$

$\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}$

$=\frac{1}{6\sqrt{\frac{36-16}{36}}}$

$=\frac{1}{6\sqrt{\frac{20}{36} }}$

$=\frac{1}{\sqrt{20}}$

$=\frac{1}{2\sqrt{5}}$

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Answer:

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Step-by-step explanation:

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then if you divide both numbers by 5 then you get 6:3

6:3 is equivalent to 2:1

sorry im a little late, hope this helped though.

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