Question

Given the two vectors \[i - j\] and \[i + 2j\] , the unit vector coplanar with the two vectors and perpendicular to the first is?
a). \[\dfrac{1}{{\sqrt 2 }}\left( {i + j} \right)\]
b). \[\dfrac{1}{{\sqrt 5 }}\left( {2i + j} \right)\]
c). \[ \pm \dfrac{1}{{\sqrt 2 }}\left( {i + j} \right)\]
d). None of these

Answer 1

Hint: Here we are asked to find the unit vector that is coplanar with the given two vectors and perpendicular with the first. First, we will find the form of the required unit vector (say c) by writing \[c = xa + yb\] since the unit vector is coplanar with the given two vectors (a and b). If the two vectors are perpendicular then their dot product equals zero using this, we can find the required unit vector.

Complete step-by-step solution:
Let us assume that the required unit vector be \[c\]. We aim to find this unit vector with the help of given data.
It is given that the vectors \[i - j\] and \[i + 2j\] are coplanar with the required unit vector \[c\]. We know that when a vector is coplanar with any two vectors then the unit vector can be written as\[c = xa + yb\].
Here the two vectors are \[a = i - j\] and \[b = i + 2j\], then the unit vector \[c\] can be written as
\[c = x\left( {i - j} \right) + y\left( {i + 2j} \right).......(1)\]
It is also given that the required unit vector is perpendicular to the first given vector that is \[c \bot a\].
We know that the dot product of the vectors perpendicular to another is zero.
Thus, \[c.a = 0\]
\[ \Rightarrow \left[ {x(i - j) + y\left( {i + 2j} \right)} \right].\left[ {i - j} \right] = 0\]
On simplifying this we get
\[ \Rightarrow \left( {x + y} \right) - \left( { - x + 2y} \right) = 0\]
\[ \Rightarrow x + y + x - 2y = 0\]
\[ \Rightarrow 2x - y = 0\]
\[ \Rightarrow y = 2x\]
On substituting the value of \[y\]in equation \[(1)\] we get
\[(1) \Rightarrow x\left( {i - j} \right) + 2x\left( {i + 2j} \right)\]
On simplifying this we get
\[ \Rightarrow c = \left( {x + 2x} \right)i + \left( { - x + 4x} \right)j\]
\[ \Rightarrow c = 3xi + 3xj.......(3)\]
Now we got the required vector but it is given that this vector is a unit vector.
So, \[\left| c \right| = 1\]\[ \Rightarrow {\left( {3x} \right)^2} + {\left( {3x} \right)^2} = 1\]
 \[ \Rightarrow 9{x^2} + 9{x^2} = 1\]
 \[ \Rightarrow 18{x^2} = 1\]
 \[ \Rightarrow {x^2} = \dfrac{1}{{18}}\]
Taking square root, we get
\[ \Rightarrow \sqrt {{x^2}} = \sqrt {\dfrac{1}{{18}}} \]
  \[ \Rightarrow x = \dfrac{1}{{3\sqrt 2 }}\]
Now substituting the above value in the equation \[(3)\] we get
\[(3) \Rightarrow c = 3\left( {\dfrac{1}{{3\sqrt 2 }}} \right)i + 3\left( {\dfrac{1}{{3\sqrt 2 }}} \right)j\]
On simplifying this we get
\[ \Rightarrow c = \dfrac{1}{{\sqrt 2 }}i + \dfrac{1}{{\sqrt 2 }}j\]
\[ \Rightarrow c = \dfrac{1}{{\sqrt 2 }}\left( {i + j} \right)\]
Thus, the required unit vector that is coplanar with \[i - j\] and \[i + 2j\], perpendicular to \[i - j\] is \[c = \dfrac{1}{{\sqrt 2 }}\left( {i + j} \right)\].
Now let us see the options, option (a) \[\dfrac{1}{{\sqrt 2 }}\left( {i + j} \right)\] is the correct option as we got the same value in our calculation above.
Option (b) \[\dfrac{1}{{\sqrt 5 }}\left( {2i + j} \right)\] is an incorrect option as we got \[\dfrac{1}{{\sqrt 2 }}\left( {i + j} \right)\] as an answer in our calculation.
Option (c) \[ \pm \dfrac{1}{{\sqrt 2 }}\left( {i + j} \right)\] is an incorrect option as we got \[\dfrac{1}{{\sqrt 2 }}\left( {i + j} \right)\] as an answer in our calculation.
Option (d) None of these is an incorrect option as we got option (a) as a correct answer.
Hence, option (a) \[\dfrac{1}{{\sqrt 2 }}\left( {i + j} \right)\] is the correct option.

Note: If \[\overrightarrow a = ({a_1},{a_2},{a_3})\] and \[\overrightarrow b = ({b_1},{b_2},{b_3})\] are any two vectors then their dot product can be written as\[\overrightarrow a .\overrightarrow b = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\]. The absolute value of a unit vector is zero. The absolute value of a vector (say \[\overrightarrow v = (x,y)\]) is\[\sqrt {{x^2} + {y^2}} \].