Question

How do you find a unit vector parallel to the x z plane and perpendicular to $\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge $ ?

Answer 1

Hint: The above problem is based on the unit vector calculation.
Vectors are the quantities which have both the magnitude and direction.
A unit vector is a vector of length 1, sometimes also called a direction vector.
Unit vector is denoted as;
$\mathop a\limits^ \wedge = \dfrac{{\mathop a\limits^ \to }}{{\left| a \right|}}$ ($\mathop a\limits^ \wedge $ is the unit vector which is equal to vector a divided by the modulus a)
With the help of above definitions we will solve the problem given to us.

Complete step-by-step solution:
Let’s discuss something about vectors first and then we will solve the problem.
Vector is the term which is used in mathematics as well as in physics and engineering, where it is defined as the object that has magnitude, direction and length. Various quantities which have both magnitude and direction are displacement, acceleration, velocity, energy, power and momentum.
A unit vector is a vector of length 1, sometimes also called a direction vector. The unit vector having the same direction as a given vector is defined by.
When a unit vector is parallel to x-z plane it means that y component must be equal to zero.
Then, the vector will become;
$x\mathop i\limits^ \wedge + z\mathop k\limits^ \wedge $ (Parallel to x z plane)
When this vector will be perpendicular to the vector$\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge $, their dot product will be equal to zero.
$ \Rightarrow (x\mathop i\limits^ \wedge + z\mathop {k)}\limits^ \wedge (\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge ) = 0$(Dot product of the two vectors is equal to zero)
$ \Rightarrow (x + 3z) = 0$ .........................1
From equation 1 we can find the value of;
$ \Rightarrow x = - 3z$ ..................2
When the unit vector is parallel to x z plane then equation of the 3- D plane will be;
$ \Rightarrow {x^2} + {z^2} = 1$
On substituting the value of x from equation 2
$ \Rightarrow {( - 3z)^2} + {z^2} = 1$
$ \Rightarrow 10{z^2} = 1$
$ \Rightarrow z = \dfrac{1}{{\sqrt {10} }}$
Then, the value of x will be:
$ \Rightarrow x = \dfrac{3}{{\sqrt {10} }}$

Then, the unit vector will be;
$ \Rightarrow \dfrac{3}{{\sqrt {10} }}\mathop i\limits^ \wedge + \dfrac{1}{{\sqrt {10} }}\mathop k\limits^ \wedge $ (Required vector)

Note: Vectors are very helpful in knowing the direction of force in which it is attempting to move the body, how the gravity exerts a force of attraction on a body to work, to calculate the motion of a body which is confined to a plane, vectors are used in engineering where the structure will sustain, to describe the force acting on a body simultaneously in 3 dimensions.