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Question: For given vectors, \(\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k\) and \(\overrightar...

For given vectors, a→=2i^−j^+2k^\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k and b→=−i^+j^−k^\overrightarrow b = - \widehat i + \widehat j - \widehat k, find the unit vector in the direction of the vector a→+b→\overrightarrow a + \overrightarrow b .

Explanation

Solution

Hint- Here, we will proceed by adding the given two vectors in order to obtain vector a→+b→\overrightarrow a + \overrightarrow b then we will divide this vector with its magnitude which will result in the unit vector in the direction of vector a→+b→\overrightarrow a + \overrightarrow b .

“Complete step-by-step answer:”
Given two vectors are a→=2i^−j^+2k^\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k and b→=−i^+j^−k^\overrightarrow b = - \widehat i + \widehat j - \widehat k
As we know that the sum of any two vectors can be obtained by adding the corresponding components of those vectors (or by adding the corresponding coefficients of i^,j^,k^\widehat i,\widehat j,\widehat k in the two vectors).
i.e., a→+b→=2i^−j^+2k^+(−i^+j^−k^) ⇒a→+b→=(2−1)i^+(−1+1)j^+(2−1)k^ ⇒a→+b→=i^−0j^+k^ ⇒a→+b→=i^+k^  \overrightarrow a + \overrightarrow b = 2\widehat i - \widehat j + 2\widehat k + \left( { - \widehat i + \widehat j - \widehat k} \right) \\\ \Rightarrow \overrightarrow a + \overrightarrow b = \left( {2 - 1} \right)\widehat i + \left( { - 1 + 1} \right)\widehat j + \left( {2 - 1} \right)\widehat k \\\ \Rightarrow \overrightarrow a + \overrightarrow b = \widehat i - 0\widehat j + \widehat k \\\ \Rightarrow \overrightarrow a + \overrightarrow b = \widehat i + \widehat k \\\
Also, we know that in order to find the unit vector in the direction of any given vector we will simply divide the given vector with the magnitude of that vector.
Magnitude for any vector x→=ai^+bj^+ck^\overrightarrow x = a\widehat i + b\widehat j + c\widehat k is given by ∣x→∣=a2+b2+c2\left| {\overrightarrow x } \right| = \sqrt {{a^2} + {b^2} + {c^2}}
Unit vector in the direction of any vector x→=ai^+bj^+ck^\overrightarrow x = a\widehat i + b\widehat j + c\widehat k is given by x^=x→∣x→∣=ai^+bj^+ck^a2+b2+c2\widehat x = \dfrac{{\overrightarrow x }}{{\left| {\overrightarrow x } \right|}} = \dfrac{{a\widehat i + b\widehat j + c\widehat k}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}
Now compare the coefficients of i^,j^,k^\widehat i,\widehat j,\widehat k in the vector a→+b→=i^+k^\overrightarrow a + \overrightarrow b = \widehat i + \widehat k with general vector x→=ai^+bj^+ck^\overrightarrow x = a\widehat i + b\widehat j + c\widehat k. From here we will get the values of a, b and c as a=1, b=0 and c=1.
The unit vector u^\widehat u in the direction of the vector a→+b→\overrightarrow a + \overrightarrow b , is given by
u^=a→+b→∣a→+b→∣=i^+k^12+02+12=i^+k^2 u^=i^2+k^2  \widehat u = \dfrac{{\overrightarrow a + \overrightarrow b }}{{\left| {\overrightarrow a + \overrightarrow b } \right|}} = \dfrac{{\widehat i + \widehat k}}{{\sqrt {{1^2} + {0^2} + {1^2}} }} = \dfrac{{\widehat i + \widehat k}}{{\sqrt 2 }} \\\ \widehat u = \dfrac{{\widehat i}}{{\sqrt 2 }} + \dfrac{{\widehat k}}{{\sqrt 2 }} \\\
Therefore, i^2+k^2\dfrac{{\widehat i}}{{\sqrt 2 }} + \dfrac{{\widehat k}}{{\sqrt 2 }} is the required unit vector in the direction of vector a→+b→\overrightarrow a + \overrightarrow b .

Note- In this particular problem,In order to find the unit vector in the direction of any given vector we will simply divide the given vector with the magnitude of that vector.we have Also, a unit vector is the vector having unit magnitude.