iv) Find a vector which is parallel to v=i-2j and has a magnitude 10.
Answers
Answer:
The vector is 2\sqrt{5} (\hat i-2\hat j)2
5
(
i
^
−2
j
^
)
Explanation:
Given vector
\vec v=\hat i-2\hat j
v
=
i
^
−2
j
^
Unit vector in the direction of vector v
\hat v=\frac{\hat i-2\hat j}{\sqrt{1^2+(-2)^2} }
v
^
=
1
2
+(−2)
2
i
^
−2
j
^
\implies \hat v=\frac{\hat i-2\hat j}{\sqrt{5} }⟹
v
^
=
5
i
^
−2
j
^
Let the required vector be a
then \vec a=|\vec a|\hat a
a
=∣
a
∣
a
^
where \hat a
a
^
is the unit vector in the direction of \vec a
a
and |\vec a|∣
a
∣ is the magnitude of \vec a
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} }
a
=10×
5
i
^
−2
j
^
\implies \vec a=2\sqrt{5} (\hat i-2\hat j)⟹
a
=2
5
(
i
^
−2
j
^
)
Thus, the required vector is 2\sqrt{5} (\hat i-2\hat j)2
5
(
i
^
−2
j
^
)