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Anvisha [2.4K]
3 years ago
14

Find a vector of magnitude 3 in the direction of v = 10i - 24k. The vector is Oi-Oj+k (Simplify your answer. Use integers or fra

ctions for any numbers in the expression).
Mathematics
1 answer:
Soloha48 [4]3 years ago
6 0

Given :

A vector v = 10i - 24k.

To Find :

A vector of magnitude 3 in the direction of v = 10i - 24k.

Solution :

Unit vector in the direction  of vector v is :

\^{v}=\dfrac{v}{|v|}\\\\\^{v}=\dfrac{10i-24k}{\sqrt{10^2+24^2}}\\\\\^{v}=\dfrac{10i-24k}{26}

Now , vector of magnitude 3 in direction v is :

r=3 \^{v}\\\\r=3\times \dfrac{ (10i-24k)}{26}\\\\r=\dfrac{3}{13} (5i-12k)

Hence , this is the required solution .

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Vanyuwa [196]

Answer:

The equation of the straight line is 4x +y = 1

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given points are (-1,5) and ( 2,-7)

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m =\frac{y_{2}-y_{1}  }{x_{2} -x_{1} }  =\frac{-7-5}{2-(-1)}  = \frac{-12}{3} = -4

slope of the line m = -4

<u><em>Step(ii):-</em></u>

The equation of the straight line passing through the point (-1,5) and having slope 'm' = -4

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y -5 = -4 x -4

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<u><em>Final answer:-</em></u>

The equation of the straight line is 4x +y = 1

<u><em>  </em></u>

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2 years ago
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Hi!

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