Question

Find the angle between the vectors {{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {A} }} = 2\widehat {\text{i}} + 3\widehat {\text{j}} - 4\widehat {\text{k}} and {{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {B} }} = 5{{\hat i}} + 2{{\hat j}} + 4{{\hat k}}.

Answer 1

Hint: The angle between two vectors can be obtained by the dot product of the two vectors. The dot product is defined as the magnitude of the component of one of the vectors with respect to the second vector. Mathematically, it can be written as-
{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {A} }}.{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {B} }} = \left| {{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {A} }}} \right|\left| {{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {B} }}} \right|cos\theta ,\text{where } {\theta } \text{ is the angle between the vectors}...(1)

Complete step-by-step answer:
We have been given two vectors {{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {A} }} = 2\widehat {\text{i}} + 3\widehat {\text{j}} - 4\widehat {\text{k}} and {{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {B} }} = 5{{\hat i}} + 2{{\hat j}} + 4{{\hat k}}. First we will find the magnitudes of vectors A and B using the formula-
$\left| {{{x\hat i}} + {{y\hat j}} + {{z\hat k}}} \right| = \sqrt {{{\text{x}}^2} + {{\text{y}}^2} + {{\text{z}}^2}}$
\begin{aligned} &\left| {{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {A} }}} \right| = \sqrt {{2^2} + {3^2} + {(-4)^2}} = \sqrt {4 + 9 + 16} = \sqrt {29} \\\ &\left| {{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {B} }}} \right| = \sqrt {{5^2} + {2^2} + {4^2}} = \sqrt {25 + 4 + 16} = \sqrt {45} \\\ \end{aligned}
Also, using the dot product formula-
{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {X} }}.{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {Y} }} = \left( {{{\text{a}}_1}{{\hat i}} + {{\text{b}}_1}{{\hat j}} + {{\text{c}}_1}{{\hat k}}} \right).\left( {{{\text{a}}_2}{{\hat i}} + {{\text{b}}_2}{{\hat j}} + {{\text{c}}_2}{{\hat k}}} \right) = {{\text{a}}_1}{{\text{a}}_2} + {{\text{b}}_1}{{\text{b}}_2} + {{\text{c}}_1}{{\text{c}}_2}
{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {A} }}.{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {B} }} = \left( {2{{\hat i}} + 3{{\hat j}} - 4{{\hat k}}} \right).\left( {5{{\hat i}} + 2{{\hat j}} + 4{{\hat k}}} \right) = 10 + 6 - 16 = 0
Substituting these values in equation (1) we get-
\begin{aligned} &{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {A} }}.{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {B} }} = \left| {{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {A} }}} \right|\left| {{{\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\rightharpoonup}}} {B} }}} \right|cos\theta \\\ &0 = \sqrt {29} \times \sqrt {45} \times cos\theta \\\ &cos;\theta = 0 \\\ {{\theta }} = {90^{\text{o}}} \\\ \end{aligned}

This is the angle between the two vectors.

Note: The result obtained in this question can be used as a general property, that is, if the dot product of any two vectors is zero, then they are perpendicular to each other. Also, in such types of questions, we can also use the formula for cross product, but it is not advisable because it is a lengthy method.