Question

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 v is pi/3

Answer 1

Answer:k=1,k=-2,k=
`8\pm 5√(3)`

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|=√(\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
`(\pi )/(3)`

`u\dot v=|u||v|cos\left ((\pi )/(3)\right )`

`|u|=√(1^2+k^2)`

`|v|=√(2^2+1^2)`

`2+k=\left ( √(1+k^2)\right )\left ( √(5)\right )cos\left ( (\pi )/(3)\right )`

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

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

k=
`8\pm 5√(3)`