Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If $\overrightarrow a ,\overrightarrow {{\text{ }}b} ,\overrightarrow {{\text{ }}c}$ are the unit vectors such that $\overrightarrow a \cdot \overrightarrow b = 0 = \overrightarrow a \cdot \overrightarrow b$ and the angle between $\overrightarrow b$ and $\overrightarrow c$ is $\dfrac{\pi }{3}$. Then find the value of $\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \left. {\overrightarrow c } \right|} \right.$

Last updated date: 18th Jun 2024
Total views: 403.8k
Views today: 10.03k
Verified
403.8k+ views
Hint: In this question we have to find the magnitude of the given vector using the given condition using unit vectors and angle. For that first we are going to find the unit vector and dot product value and then using a given angle we can find the required cross product value of the given vector.

From the question, $\overrightarrow a ,\overrightarrow b ,\overrightarrow c$ are unit vector
Then the magnitude value of unit vector is $1$
$\therefore \left| {\overrightarrow a } \right| = 1$, $\left| {\overrightarrow b } \right| = 1$,$\left| {\overrightarrow c } \right| = 1$
We know that if the dot product of two vector is 0 then the two vectors are said to be perpendicular,
$\therefore \overrightarrow a .\overrightarrow b = 0$ then $\overrightarrow a$ is perpendicular to $\overrightarrow b$
$\therefore \overrightarrow a .\overrightarrow c = 0$ then $\overrightarrow a$ is perpendicular to $\overrightarrow c$
So, $\overrightarrow a$ is perpendicular to both $\overrightarrow b$ and $\overrightarrow c$ then $\overrightarrow b$ and $\overrightarrow c$ are parallel, then we obtain that $\overrightarrow a$ is parallel to $\overrightarrow b \times \overrightarrow c$.
That is, $\overrightarrow a$ is parallel to $\overrightarrow b \times \overrightarrow c$
Thus, we obtain $\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \left. {\overrightarrow c } \right|} \right.$
Now take $a$ common, we get
$\Rightarrow \left| {\overrightarrow a \times (\overrightarrow b - \left. {\overrightarrow c )} \right|} \right. - - - - - - - - - - - (1)$
We know that if $\overrightarrow b$ and $\overrightarrow c$ are parallel, then
$\Rightarrow \overrightarrow b = \lambda \overrightarrow c$
Now substitute $\overrightarrow b$ value in equation (1), we get
$\Rightarrow \left| {\overrightarrow a \times (\lambda \overrightarrow c - \left. {\overrightarrow c )} \right|} \right.$
Now taking $\overrightarrow c$ common, we get
$\Rightarrow \left| {\overrightarrow a \times (\lambda - \left. {1)\overrightarrow c } \right|} \right.$
We can separate ${\text{(}}\lambda {\text{ - 1)}}$ from the modulus, we get
$\Rightarrow {\text{(}}\lambda {\text{ - 1)}}\left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. - - - - - - - - - - - - (2)$
We know that,
$\Rightarrow \left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \theta - - - - - - - - - - (3)$
Given that the angle between $\overrightarrow a$ and $\overrightarrow c$ is $\dfrac{\pi }{3}$ because $\overrightarrow a$ and $\overrightarrow c$ is perpendicular and substitute the angle in equation (3), we get
$\Rightarrow \left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \dfrac{\pi }{2}$
Since $\left| {\overrightarrow a } \right| = 1$, $\left| {\overrightarrow c } \right| = 1$, then
$\Rightarrow \left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. = 1 \cdot 1.\sin {90^ \circ }$
$\Rightarrow \left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. = \sin {60^ \circ } - - - - - - - - - - (4)$
We know that, $\sin {90^ \circ } = 1$ substitute $\sin {60^ \circ }$ value in equation (4), we get
$\Rightarrow \left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. = 1 - - - - - - - - - - (5)$
Now, Substitute equation (5) in (2)
$\Rightarrow \left( {\lambda {\text{ - 1}}} \right)\left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right.$
$\Rightarrow \left( {\lambda - 1} \right) \cdot 1$
$\Rightarrow \left( {\lambda - 1} \right) - - - - - - - - - - - - - - - (6)$
Similarly, now take $\overrightarrow a$ is parallel to $\overrightarrow b \times \overrightarrow c$,
We know that if $\overrightarrow b$ and $\overrightarrow c$ are parallel, then
$\Rightarrow \overrightarrow b = \lambda \overrightarrow c$
Similarly $\overrightarrow a$ is parallel to $\overrightarrow b \times \overrightarrow c$
$\Rightarrow \overrightarrow a = \lambda (\overrightarrow b \times \overrightarrow c )$
$\Rightarrow \left| {\overrightarrow a } \right| = \lambda \left| {(\overrightarrow b \times \overrightarrow c )} \right|$
We know that $\overrightarrow a ,{\text{ }}\overrightarrow b ,{\text{ }}\overrightarrow c$ are unit vectors
$\Rightarrow \left| {\overrightarrow a } \right| = \lambda \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin {60^ \circ }$
Given that $\left| {\overrightarrow a } \right| = 1$, $\left| {\overrightarrow b } \right| = 1$, $\left| {\overrightarrow c } \right| = 1$and $\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$
By substituting, we get
$\Rightarrow 1 = \lambda \cdot 1 \cdot 1 \cdot \dfrac{{\sqrt 3 }}{2}$
Now we get $\lambda$ value,
$\Rightarrow \lambda = \dfrac{2}{{\sqrt 3 }}$
Now substitute $\lambda$ value in equation (6), we get
$\Rightarrow \left( {\lambda - 1} \right) - - - - - - - - - - - - - - - (6)$
$\Rightarrow \left( {\dfrac{2}{{\sqrt 3 }} - 1} \right)$
Multiply and divide by $\sqrt 3$ we get,
$\Rightarrow \left( {\dfrac{2}{{\sqrt 3 }} - \dfrac{{\sqrt 3 }}{{\sqrt 3 }}} \right)$
Denominator for the both terms are equal, we get,
$\Rightarrow \left( {\dfrac{{2 - \sqrt 3 }}{{\sqrt 3 }}} \right)$

$\therefore$ Thus the value $\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \left. {\overrightarrow c } \right|} \right.$ is $\dfrac{{2 - \sqrt 3 }}{{\sqrt 3 }}$

Note: This kind of problem has to be derived in a substitution way because the question gives ideas to solve in the manner of dot product as well as cross product. The basic vector calculation and algebraic calculation take place. And also we need to know some trigonometric ratio values and fraction addition.