Question

Vector a and c are unit vectors and $\left| \overrightarrow{b} \right|\text{ = 4}$ . If angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is ${{\cos }^{-1}}\left( \dfrac{1}{4} \right)$ and \[\overrightarrow{a}\times \overrightarrow{b}\text{ = 2}\overrightarrow{a}\times \overrightarrow{c}\] , then $\overrightarrow{b}\text{ = }\lambda \overrightarrow{a}\text{ + 2}\overrightarrow{c}$. Find the value of $\lambda $ .
A. $\pm \text{ }\dfrac{1}{4}$
B. $\pm \text{ }\dfrac{1}{2}$
C. $\pm 4$
D. None of the above

Answer 1

Hint: Use the equation \[\overrightarrow{a}\times \overrightarrow{b}\text{ = 2}\overrightarrow{a}\times \overrightarrow{c}\] to obtain the relation $\overrightarrow{b}\text{ = }\lambda \overrightarrow{a}\text{ + 2}\overrightarrow{c}$. Adjust the relation accordingly so as to use the conditions given in the question. Take modulus on both sides of the equation and thus find the value of $\lambda $.

Complete step-by-step answer:
According to the question, \[\overrightarrow{a}\times \overrightarrow{b}\text{ = 2}\overrightarrow{a}\times \overrightarrow{c}\]
Now, this can be adjusted in the form,
\[\begin{align}
  & \left( \overrightarrow{a}\times \overrightarrow{b} \right)\text{ }-\text{ }\left( 2\overrightarrow{a}\times \overrightarrow{c} \right)\text{ = }\overrightarrow{0} \\
 & \Rightarrow \text{ }\overrightarrow{a}\times \left( \overrightarrow{b}\text{ }-\text{ 2}\overrightarrow{c} \right)\text{ = }\overrightarrow{0} \\
\end{align}\]
We know, the cross product of two vectors is a zero vector, if and only if the two vectors are collinear.
Thus, we can express this in the form,
\[\left( \overrightarrow{b}-\text{2}\overrightarrow{c} \right)\text{ = }\lambda \overrightarrow{a}\text{ }.....\left( \text{i} \right)\]
Here, $\lambda $is the linear coefficient factor between the two vectors $\overrightarrow{b}-2\overrightarrow{c}$ and $\overrightarrow{a}$.
Equation (i) can be readjusted as $\overrightarrow{b}\text{ = }\lambda \overrightarrow{a}\text{ + 2}\overrightarrow{c}$ . Thus, we get the relation between the three vectors as given in the question.
Taking modulus on both sides of the equation (i), we can write,
$\begin{align}
  & \left| \overrightarrow{b}\text{ }-\text{ 2}\overrightarrow{c} \right|\text{ = }\left| \lambda \overrightarrow{a} \right| \\
 & \Rightarrow \text{ }{{\left| \overrightarrow{b}\text{ }-\text{ 2}\overrightarrow{c} \right|}^{2}}\text{ = }{{\left| \lambda \overrightarrow{a} \right|}^{2}}\text{ }\left( \text{Squaring both sides as they are scalar quantities} \right) \\
 & \Rightarrow \text{ }{{\left| \overrightarrow{b} \right|}^{2}}\text{ + }{{\left| 2\overrightarrow{c} \right|}^{2}}\text{ }-\text{ 2}\cdot \overrightarrow{b}\cdot 2\overrightarrow{c}\text{ = }{{\lambda }^{2}}{{\left| \overrightarrow{a} \right|}^{2}} \\
 & \Rightarrow \text{ }{{\left| \overrightarrow{b} \right|}^{2}}\text{ + 4}\cdot {{\left| \overrightarrow{c} \right|}^{2}}\text{ }-\text{ 4}\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\cos \left( {{\cos }^{-1}}\dfrac{1}{4} \right)\text{ = }{{\lambda }^{2}}\cdot {{\left| \overrightarrow{a} \right|}^{2}}\text{ }\left( \text{Angle between }b\text{ and }c\text{ is }{{\cos }^{-1}}\dfrac{1}{4}\text{ } \right) \\
 & \Rightarrow \text{ }{{\text{4}}^{2}}\text{ + 4}\cdot {{\text{1}}^{2}}\text{ }-\text{ 4}\cdot \text{4}\cdot \text{1}\cdot \dfrac{1}{4}\text{ = }{{\lambda }^{2}}\cdot {{1}^{2}}\text{ }\left( \text{Putting the values }\left| a \right|=1,\left| b \right|=4,\left| c \right|=1 \right) \\
 & \Rightarrow \text{ 16 + 4 }-\text{ 4 = }{{\lambda }^{2}} \\
 & \Rightarrow \text{ }{{\lambda }^{2}}\text{ = 16} \\
 & \therefore \text{ }\lambda \text{ = }\pm \text{4} \\
\end{align}$
Thus, we obtain the value of $\lambda $ to be $\pm 4$
Hence, the correct answer is option C.

Note: The linear relation between the three vectors as given in the question needed to be readjusted to equation (i) and then taken modulus, because we are only given the angle between the vectors b and c. So, we have to keep b and c on the same side of the equation. If modulus was taken on the relation as given in the question, we would have needed the angle between a and c to proceed further and find the value of $\lambda $.