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Find the value of ${{\lambda }}$ in unit vector ${{0}}{{.4}}\;{{\hat i}}\;{{ + 0}}{{.8}}\;{{\hat j}}\;{{ + }}\;{{\lambda }}\;{{\hat k}}$

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Last updated date: 17th Apr 2024
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Answer
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Hint: A unit vector is that vector which have magnitude $1.$ so we will put the magnitude of this vector equal to $1$ and then calculate value of ${{\lambda }}$ , A unit vector can be in any direction. Unit vectors are helpful to determine the base form of a vector space. Every vector in a given space can be expressed as a linear combination of unit vectors.

Formula used:
$\left| {{{\vec A}}} \right|\;{{ = }}\sqrt {{{{A}}_{{x}}}^{{2}}{{ + }}{{{A}}_{{y}}}^{{2}}{{ + }}{{{A}}_{{z}}}^{{2}}} $
Where $\left| {{{\vec A}}} \right|$ is any vector and ${{{A}}_{{{x}}\;{{,}}\;}}{{{A}}_{{{y}}\;{{,}}}}{{{A}}_{{z}}}$ are its components along x, y, z directions respectively and $\left| {{{\vec A}}} \right|$ is the magnitude of this vector.

Complete step by step solution:
The given vector is $\dfrac{{{4}}}{{{{10}}}}{{\hat i}}\;{{ + }}\,\dfrac{{{8}}}{{{{10}}}}{{\hat j}}\;{{ + }}\;{{\lambda \hat k}}$
As, the vector is unit vector so, its magnitude will be equal to $1$ i.e.
$\sqrt {{{\left( {\dfrac{{{4}}}{{{{10}}}}} \right)}^{{2}}}{{ + }}{{\left( {\dfrac{{{8}}}{{{{10}}}}} \right)}^{{2}}}{{ + }}{{{\lambda }}^{{2}}}} {{ = 1}}$
$ \Rightarrow \;\dfrac{{{{16}}}}{{{{100}}}}{{ + }}\dfrac{{{{64}}}}{{{{100}}}}{{ + }}{{{\lambda }}^{{2}}}\;{{ = }}\;{{1}}$
$ \Rightarrow \dfrac{{{{100}}}}{{{{100}}}}{{ + }}{{{\lambda }}^{{2}}}\;{{ = }}\;{{1}}$
$ \Rightarrow {{\lambda }}\;{{ = }}\;{{0}}$
To find a unit vector with the same direction as a given vector, we divide the vector by its magnitude.
Any vector can be converted into a unit vector by dividing it by the magnitude of the given vector. The dot product for any two unit vectors is a scalar quantity whereas the cross product of any two arbitrary unit vectors results in a third vector orthogonal to both of them.

Note: For such a question always put the magnitude equal to $1.$ Normal vector is a vector which is perpendicular to the surface at a given point. They are also called “normal,” to a surface is a vector. When normals are estimated on any closed surfaces, the normal pointing towards the interior of the surface and normal pointing outward are usually discovered.