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2 years ago

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Please help me!!!!!!

2 answers:

Andreas93 [3]2 years ago

6 0

Answer:

Step-by-step explanation:

kvasek [131]2 years ago

4 0

2 would be the correct answer ;-)))) (1 on top?)

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Let u = (1, k) and v = (2, 1). Find k such that The distance between u and v is 1 u and v are orthogonal The angle between u and

SVETLANKA909090 [29]

Answer:k=1,k=-2,k= $8\pm 5\sqrt{3}$

Step-by-step explanation:

Given two vectors

$u=1\hat{i}+k\hat{j}$

$v=2\hat{i}+1\hat{j}$

$\left ( i\right )$ Distance between them is given by

$|u-v|=\sqrt{\left ( 2-1\right )^2+\left ( 1-k\right )^2}=1$

squaring both side

$1^{2}+\left ( 1-k\right )^2=1$

$k^2-2k+1=0$

$\left ( k-1\right )^2=0$

k=1

$\left ( ii\right )$

angle between u and v is 90 i.e. orthogonal

$u\dot v=0$

$\left ( 1\hat{i}+k\hat{j}\right )\dot \left ( 2\hat{i}+1\hat{j}\right )$ =0

2+k=0

k=-2

$\left ( iii\right )$

angle between u & v is $\frac{\pi }{3}$

$u\dot v=|u||v|cos\left (\frac{\pi }{3}\right )$

$|u|=\sqrt{1^2+k^2}$

$|v|=\sqrt{2^2+1^2}$

$2+k=\left ( \sqrt{1+k^2}\right )\left ( \sqrt{5}\right )cos\left ( \frac{\pi }{3}\right )$

$\left ( 4+2\right )^2=\left ( 1+k^2\right )5$

$k^2-16k-11$ =0

k= $8\pm 5\sqrt{3}$

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We must solve

$8 - \frac{7}{10}x = 6 - \frac{1}{5}x \\ or \\ 80 - 7x = 60 - 2x \\ now \\ 20 = 5x \: so \: x = 4 \\$
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Helga [31]

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15 may be

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lapo4ka [179]

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The roots of x^2-6x-16=0 <br> {-16,6}

LenaWriter [7]

Answer:

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2 years ago