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Question: Determine a unit vector which is perpendicular to both \[A = 2i + j + k\] and \(B = i - j + 2k\) ?...

Determine a unit vector which is perpendicular to both A=2i+j+kA = 2i + j + k and B=i−j+2kB = i - j + 2k ?

Explanation

Solution

In cross product (or vector product) of two nonzero vectors a⃗\vec a and b⃗\vec b , the resultant vector is perpendicular to both vectors a⃗\vec a and b⃗\vec b
So we got the hint, to find a vector perpendicular to two nonzero vectors a⃗\vec a and b⃗\vec b , we have to find the cross product of those two vectors.
Remember that the resultant vector may or may not be a unit vector.
Unit vectors are those vectors whose magnitude is 1.
Therefore, find the unit vector by dividing the vector by its magnitude.

Complete step by step answer:
Step-1: Find the cross product of a⃗\vec a and b⃗\vec b .
a^×b^\hat a \times \hat b is the determinant of the matrix \left( {\begin{array}{*{20}{c}} i&j;&k; \\\ 2&1&1 \\\ 1&{ - 1}&2 \end{array}} \right)
\Rightarrow \hat a \times \hat b = \left( {\begin{array}{*{20}{c}} i&j;&k; \\\ 2&1&1 \\\ 1&{ - 1}&2 \end{array}} \right) \\\ \Rightarrow \hat i(1 \times 2 - ( - 1) \times 1) - \hat j(2 \times 2 - 1 \times 1) + \hat k(2 \times ( - 1) - 1 \times 1) \\\ \Rightarrow 3\hat i - 3\hat j - 3\hat k \\\
Let c⃗=a^×b^\vec c = \hat a \times \hat b
Step-2: Find the unit vector c^\hat c
c⃗=3i^−3j^−3k^\vec c = 3\hat i - 3\hat j - 3\hat k
Magnitude of c⃗\vec c
∣c⃗∣=(3)2+(−3)2+(−3)2 ⇒∣c⃗∣=27 ⇒∣c⃗∣=33  \left| {\vec c} \right| = \sqrt {{{(3)}^2} + {{( - 3)}^2} + {{( - 3)}^2}} \\\ \Rightarrow \left| {\vec c} \right| = \sqrt {27} \\\ \Rightarrow \left| {\vec c} \right| = 3\sqrt 3 \\\
And we know that Unit vector is c^=c⃗∣c∣\hat c = \dfrac{{\vec c}}{{\left| c \right|}}
∴c^=3i^−3j^−3k^33\therefore \hat c = \dfrac{{3\hat i - 3\hat j - 3\hat k}}{{3\sqrt 3 }}
So, c^=i^3−j^3−k^3\hat c = \dfrac{{\hat i}}{{\sqrt 3 }} - \dfrac{{\hat j}}{{\sqrt 3 }} - \dfrac{{\hat k}}{{\sqrt 3 }}
Hence the Unit vector perpendicular to both a⃗\vec a and b⃗\vec b is i^3−j^3−k^3\dfrac{{\hat i}}{{\sqrt 3 }} - \dfrac{{\hat j}}{{\sqrt 3 }} - \dfrac{{\hat k}}{{\sqrt 3 }}

Note: The unit vector i^\hat i is along the direction of the x-axis, the unit vector j^\hat j is along the direction of the y-axis, and the unit vector k^\hat k is along the direction of the z-axis. Thus, i^\hat i , j^\hat j and k^\hat k are unit vectors mutually perpendicular to each other.