Question

find a vector which is parallel to v=i-2j and has a magnitude 10​

Answer 1

Answer:

The vector is $2\sqrt{5} (\hat i-2\hat j)$

Explanation:

Given vector

$\vec v=\hat i-2\hat j$

Unit vector in the direction of vector v

$\hat v=\frac{\hat i-2\hat j}{\sqrt{1^2+(-2)^2} }$

$\implies \hat v=\frac{\hat i-2\hat j}{\sqrt{5} }$

Let the required vector be a

then $\vec a=|\vec a|\hat a$

where $\hat a$ is the unit vector in the direction of $\vec a$ and $|\vec a|$ is the magnitude of $\vec a$

Since v and a vectors are parallel the unit vectors will be same

Therefore,

$\vec a=10\times \frac{\hat i-2\hat j}{\sqrt{5} }$

$\implies \vec a=2\sqrt{5} (\hat i-2\hat j)$

Thus, the required vector is $2\sqrt{5} (\hat i-2\hat j)$

Answer 2

Answer:

v=i-2j. .x=10

uv= v/v

v/v = x/x

x=x(v/v)

10(i-2j/√5)

(V=√1*1 +2*2/√5)

x=(10/√5)i-(20/√5)j

This Answer is Perfect,