Question

The unit vector in ZOX- plane and making angle ${45^ \circ }$ and ${60^ \circ }$, respectively with $a = 2i + 2j - k$ and $b = 0i + j - k$ is?
A) $\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)i + \left( {\dfrac{1}{{\sqrt 2 }}} \right)k$
B) $\left( {\dfrac{1}{{\sqrt 2 }}} \right)i - \left( {\dfrac{1}{{\sqrt 2 }}} \right)k$
C) $\left( {\dfrac{1}{{3\sqrt 2 }}} \right)i + \left( {\dfrac{4}{{3\sqrt 2 }}} \right)j + \left( {\dfrac{1}{{3\sqrt 2 }}} \right)k$
D) None of the above

Answer 1

Hint: In this question it is given that the required vector is a unit vector. It means that the magnitude of our required vector is one. Also, the required vector is making some angle with the other vector given in the question. So, we can use that formula for finding the angle between two vectors to solve this question.
Formula used: $\cos \theta = \dfrac{{m.n}}{{|m||n|}}$

Complete step by step answer:
Let, the required vector be $\overrightarrow r = x\hat i + y\hat j + z\hat k$.
But, in the above question it is given that the vector is in the x-z plane.
Therefore, the y-component of the vector is $0.$
$\overrightarrow r = x\hat i + z\hat k$
In the above question it is given that the required vector is a unit vector.
Therefore, the magnitude of the required vector is $1.$
$|\overrightarrow {r|} = \sqrt {{x^2} + {z^2}} $
$1 = \sqrt {{x^2} + {z^2}} $
${x^2} + {z^2} = 1$
Now,
Vector $\overrightarrow r $ makes an angle of ${45^ \circ }$ with vector $\overrightarrow a $.
Using the formula, $\cos \theta = \dfrac{{m.n}}{{|m||n|}}$
$\cos {45^ \circ } = \dfrac{{\left( {\overrightarrow r } \right).\left( {\overrightarrow a } \right)}}{{|\overrightarrow r ||\overrightarrow a |}} = \dfrac{{\left( {x\hat i + z\hat k} \right).\left( {2\hat i + 2\hat j - \hat k} \right)}}{{1 \times \sqrt {{2^2} + {2^2} + 1} }}$
We know that, $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ and $|\overrightarrow {r| = 1} $
Using dot product in numerator in RHS
$\dfrac{1}{{\sqrt 2 }} = \dfrac{{2x - z}}{{\sqrt 9 }}$
$\dfrac{3}{{\sqrt 2 }} = 2x - z - - - - - \left( 1 \right)$
Also, vector $\overrightarrow r $ makes an angle of ${60^ \circ }$with $\overrightarrow {b.} $
Again, using the formula
$\cos {60^ \circ } = \dfrac{{\left( {x\hat i + z\hat k} \right).\left( {0\hat i + \hat j - \hat k} \right)}}{{1 \times \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2}} }}$
Using dot product
$\dfrac{1}{2} = \dfrac{{ - z}}{{\sqrt 2 }}$
$z = \dfrac{{ - 1}}{{\sqrt 2 }}$
Now, putting the value of z in equation $\left( 1 \right)$
$\dfrac{3}{{\sqrt 2 }} = 2x - \left( { - \dfrac{1}{{\sqrt 2 }}} \right)$
$\dfrac{3}{{\sqrt 2 }} = 2x + \dfrac{1}{{\sqrt 2 }}$
On transposing, we get
$\dfrac{3}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 2 }} = 2x$
$\sqrt 2 = 2x$
$x = \dfrac{1}{{\sqrt 2 }}$
Now putting the value, $z = \dfrac{{ - 1}}{{\sqrt 2 }}$ and $x = \dfrac{1}{{\sqrt 2 }}$.
Therefore, the required vector is $\dfrac{1}{{\sqrt 2 }}\hat i - \dfrac{1}{{\sqrt 2 }}\hat k$.
Hence, the correct option is option (B).

Note:
There are several concepts used in this question like dot product and the formula of angle between two vectors. Concept of unit vectors is also used in this question, which is a very important concept of vectors. These concepts should be kept in mind while solving the questions of vectors.