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Question: Find a unit vector in the direction of the given vector of \[v = (5, - 12)\]?...

Find a unit vector in the direction of the given vector of v=(5,−12)v = (5, - 12)?

Explanation

Solution

Hint : In the given question, we are given coordinates of a point and we are asked to find a unit vector in the direction of the point. So, we first find the position vector of the coordinates given to us. Then, we find the unit vector in the direction of the given vector by dividing the vector by the magnitude of the same vector.

Complete step by step solution:
The position vector of the given coordinates v is v⃗=5i^−12j^\vec v = 5\hat i - 12\hat j . Hence, we now have to find the unit vector in the direction of the above vector.
So, A unit vector in the direction of a given vector, say a⃗\vec a can be found as a^=a⃗∣a∣\hat a = \dfrac{{\vec a}}{{\left| a \right|}} .
Now, Let a⃗=xi^+yj^+zk^\vec a = x\hat i + y\hat j + z\hat k .
Then, ∣a∣=x2+y2+z2\left| a \right| = \sqrt {{x^2} + {y^2} + {z^2}}
So, we get the unit vector in the direction of the vector a⃗=xi^+yj^+zk^\vec a = x\hat i + y\hat j + z\hat k by dividing the vector by it’s magnitude as: a^=(xi^+yj^+zk^x2+y2+z2)\hat a = \left( {\dfrac{{x\hat i + y\hat j + z\hat k}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) .
For the given vector, v⃗=5i^−12j^\vec v = 5\hat i - 12\hat j , we get the magnitude as ∣v⃗∣=(5)2+(−12)2\left| {\vec v} \right| = \sqrt {{{(5)}^2} + {{( - 12)}^2}} .
Hence, ∣v⃗∣=25+144\left| {\vec v} \right| = \sqrt {25 + 144}
⇒∣v⃗∣=169\Rightarrow \left| {\vec v} \right| = \sqrt {169}
⇒∣v⃗∣=13\Rightarrow \left| {\vec v} \right| = 13
So, v^=5i^−12j^(5)2+(−12)2\hat v = \dfrac{{5\hat i - 12\hat j}}{{\sqrt {{{(5)}^2} + {{( - 12)}^2}} }}
⇒v^=5i^−12j^169\Rightarrow \hat v = \dfrac{{5\hat i - 12\hat j}}{{\sqrt {169} }}
⇒v^=(5i^−12j^13)\Rightarrow \hat v = \left( {\dfrac{{5\hat i - 12\hat j}}{{13}}} \right)
⇒v^=(513i^−1213j^)\Rightarrow \hat v = \left( {\dfrac{5}{{13}}\hat i - \dfrac{{12}}{{13}}\hat j} \right)
Hence, the unit vector in the direction of the given vector of v=(5,−12)v = (5, - 12) is v^=(513i^−1213j^)\hat v = \left( {\dfrac{5}{{13}}\hat i - \dfrac{{12}}{{13}}\hat j} \right) .
So, the correct answer is “ v^=(513i^−1213j^)\hat v = \left( {\dfrac{5}{{13}}\hat i - \dfrac{{12}}{{13}}\hat j} \right) ”.

Note : The position vector is the vector joining the origin to the point given to us. The position vector is obtained by taking into consideration the definition of the position vector and the coordinates of the point given to us. The magnitude of the vector represents the length of the vector and is the real number.