Question

If \[\lambda (3i+2j-6k)\]is a unit vector, then the values of \[\lambda \]are?

Answer 1

Hint:To get the solution of this problem we have to use the formula of unit vectors. For that, firstly calculate the magnitude of the given vector by using the magnitude formula and then compare from the question to get the value of \[\lambda \]. By using these steps you can now find the final answer.

Complete step-by-step solution:
A vector is a physical quantity that has both direction and magnitude. There are different types of vectors: unit vector, position vector, zero vector, equal vector and many more. Let us discuss unit vectors which we have to use in this question.
Unit vector means it has both the quantity i.e. magnitude and direction. For the unit vector the magnitude is always equal to\[1\]. Unit vector is also called by the name direction vector. Any of the vectors can become a unit vector by dividing the given vector by its magnitude itself. The unit vector is represented by the symbol \[{{''}^{\wedge }}''\] this symbol is known as “cap” or “hat”.
The formula to find the unit vector is given by
\[\overset{\wedge }{\mathop{a}}\ =\dfrac{{\vec{a}}}{\left| {\vec{a}} \right|} \]
where \[\vec{a}\] represents the given vector and \[\left| {\vec{a}} \right|\] represents the magnitude of the given vector.
The dot product of any two vectors is the product of magnitude of the two vectors and the cosine of the angle between them; it always results in the scalar quantity whereas cross product of magnitude of the two vectors and the sine of the angle between them. Cross product always results in vector quantity.
Magnitude is the length of the vector. Magnitude of any vector is calculated by summation of squares of all the components and then square root to get the final answer.
In this question the given vector is \[\vec{a}=3i+2j-6k\]. It is given that a vector is a unit vector. So, by the formula of unit vector, we get
\[\begin{align}
  & \Rightarrow \overset{\wedge }{\mathop{a}}\,=\dfrac{3i+2j-6k}{\sqrt{{{(3)}^{2}}+{{(2)}^{2}}+{{(-6)}^{2}}}} \\
 & \Rightarrow \overset{\wedge }{\mathop{a}}\,=\dfrac{3i+2j-6k}{\sqrt{9+4+36}} \\
 & \Rightarrow \overset{\wedge }{\mathop{a}}\,=\dfrac{3i+2j-6k}{\sqrt{49}} \\
\end{align}\]
\[\Rightarrow \overset{\wedge }{\mathop{a}}\,=\dfrac{3i+2j-6k}{\pm 7}\]
By comparing from the given question, we get the value of \[\lambda \]as
\[\Rightarrow \lambda =\pm \dfrac{1}{7}\]
So we can conclude that the final answer is \[\pm \dfrac{1}{7}\].

Note:Vectors in physics are used to represent the quantities force, torque, acceleration, displacement, velocity and so on. The dot product is always equal to \[0\]if both the vectors are perpendicular whereas the cross product is always equal to \[0\] if both the vectors are in the same direction.