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2 years ago

15

01101 + 01001 please help me

1 answer:

Marrrta [24]2 years ago

6 0

Answer: 2102

Step-by-step explanation:

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3 years ago

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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2 years ago