Question

iv) Find a vector which is parallel to v=i-2;
and has a magnitude 10​

Answer 1

Answer:

The vector is 2\sqrt{5} (\hat i-2\hat j)25(i^−2j^)

Explanation:

Given vector

\vec v=\hat i-2\hat jv=i^−2j^

Unit vector in the direction of vector v

\hat v=\frac{\hat i-2\hat j}{\sqrt{1^2+(-2)^2} }v^=12+(−2)2i^−2j^

\implies \hat v=\frac{\hat i-2\hat j}{\sqrt{5} }⟹v^=5i^−2j^

Let the required vector be a

then \vec a=|\vec a|\hat aa=∣a∣a^

where \hat aa^ is the unit vector in the direction of \vec aa and |\vec a|∣a∣ is the magnitude of \vec aa

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} }a=10×5i^−2j^

\implies \vec a=2\sqrt{5} (\hat i-2\hat j)⟹a=25(i^−2j^)

Thus, the required vector is 2\sqrt{5} (\hat i-2\hat j)25(i^−2j^)