Question

Find the value of $\left| \left( \overrightarrow{a}\times \overrightarrow{b} \right) \right|$, if $\left| \overrightarrow{a} \right|=4$, $\left| \overrightarrow{b} \right|=5$ and $\left( \overrightarrow{a}.\overrightarrow{b} \right)=-6$
(A) 18
(B) $\sqrt{364}$
(C) 19
(D) $\sqrt{399}$

Answer 1

We solve this question by first using the formula for the dot product, $\left( \overrightarrow{a}.\overrightarrow{b} \right)=\left| \overrightarrow{a} \right|\times \left| \overrightarrow{b} \right|\times \cos \theta$ and substitute the given values of $\left| \overrightarrow{a} \right|$ and $\left| \overrightarrow{b} \right|$, and find the value of $\sin \theta$. Then we use the formula for the magnitude of cross product, $\left( \left| \overrightarrow{a}\times \overrightarrow{b} \right| \right)=\left| \overrightarrow{a} \right|\times \left| \overrightarrow{b} \right|\times \sin \theta$ and substitute the given values of $\left| \overrightarrow{a} \right|$ and $\left| \overrightarrow{b} \right|$, and find the value in terms of $\cos \theta$. Then we use the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, and use the value of $\sin \theta$ to find the value of $\cos \theta$ and substitute in the formula for the magnitude of the cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$, that is the required value.

Complete step by step answer:
We are given that the magnitudes of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are 4 and 5 respectively, that is
$\begin{aligned} & \Rightarrow \left| \overrightarrow{a} \right|=4 \\\ & \Rightarrow \left| \overrightarrow{b} \right|=5 \\\ \end{aligned}$
We are also given that the dot product of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is -6, that is $\left( \overrightarrow{a}.\overrightarrow{b} \right)=-6$.
Now let us consider the formula for the dot product of any two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.
$\left( \overrightarrow{a}.\overrightarrow{b} \right)=\left| \overrightarrow{a} \right|\times \left| \overrightarrow{b} \right|\times \cos \theta$
, where $\theta$ is the angle between those vectors.
Using the above formula for the dot product and substituting the above values of the magnitudes and dot product, we get

& \Rightarrow \left( \overrightarrow{a}.\overrightarrow{b} \right)=\left| \overrightarrow{a} \right|\times \left| \overrightarrow{b} \right|\times \cos \theta \\\ & \Rightarrow -6=4\times 5\times \cos \theta \\\ & \Rightarrow -6=20\times \cos \theta \\\ & \Rightarrow \cos \theta =\dfrac{-6}{20} \\\ & \Rightarrow \cos \theta =\dfrac{-3}{10}..........\left( 1 \right) \\\ \end{aligned}$$ We need to find the value of $\left| \left( \overrightarrow{a}\times \overrightarrow{b} \right) \right|$. Now, let us consider the formula for the magnitude of the cross product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$. $$\left( \left| \overrightarrow{a}\times \overrightarrow{b} \right| \right)=\left| \overrightarrow{a} \right|\times \left| \overrightarrow{b} \right|\times \sin \theta $$ , where $\theta $ is the angle between those vectors. Using the above formula for the dot product and substituting the above values of the magnitudes and dot product, we get $$\begin{aligned} & \Rightarrow \left| \overrightarrow{a}\times \overrightarrow{b} \right|=\left| \overrightarrow{a} \right|\times \left| \overrightarrow{b} \right|\times \sin \theta \\\ & \Rightarrow \left| \overrightarrow{a}\times \overrightarrow{b} \right|=4\times 5\times \sin \theta \\\ & \Rightarrow \left| \overrightarrow{a}\times \overrightarrow{b} \right|=20\sin \theta .........\left( 2 \right) \\\ \end{aligned}$$ So, to find the value of $$\left| \overrightarrow{a}\times \overrightarrow{b} \right|$$, we need to find the value of $\sin \theta $. From equation (1), we know the value of $\cos \theta $. So, now let us consider the trigonometric identity, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ So, using this identity and the equation (1), we get $\begin{aligned} & \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ & \Rightarrow {{\sin }^{2}}\theta +{{\left( -\dfrac{3}{10} \right)}^{2}}=1 \\\ & \Rightarrow {{\sin }^{2}}\theta +\,\dfrac{9}{100}=1 \\\ & \Rightarrow {{\sin }^{2}}\theta =1-\dfrac{9}{100}=\dfrac{91}{100} \\\ & \Rightarrow \sin \theta =\dfrac{\sqrt{91}}{10} \\\ \end{aligned}$ Substituting this value in the equation (2) we get, $$\begin{aligned} & \Rightarrow \left| \overrightarrow{a}\times \overrightarrow{b} \right|=20\times \dfrac{\sqrt{91}}{10} \\\ & \Rightarrow \left| \overrightarrow{a}\times \overrightarrow{b} \right|=2\sqrt{91} \\\ & \Rightarrow \left| \overrightarrow{a}\times \overrightarrow{b} \right|=\sqrt{4\times 91} \\\ & \Rightarrow \left| \overrightarrow{a}\times \overrightarrow{b} \right|=\sqrt{364} \\\ \end{aligned}$$ So, we get that the value of $\left| \left( \overrightarrow{a}\times \overrightarrow{b} \right) \right|$ is $$\sqrt{364}$$. **So, the correct answer is “Option B”.** **Note:** The general mistake one makes while solving this type of questions is one might take the formulas for the dot product and cross product of the vectors in reverse as, $$\begin{aligned} & \Rightarrow \left( \left| \overrightarrow{a}\times \overrightarrow{b} \right| \right)=\left| \overrightarrow{a} \right|\times \left| \overrightarrow{b} \right|\times \cos \theta \\\ & \Rightarrow \left( \overrightarrow{a}.\overrightarrow{b} \right)=\left| \overrightarrow{a} \right|\times \left| \overrightarrow{b} \right|\times \sin \theta \\\ \end{aligned}$$ But the actual formulas are $$\begin{aligned} & \Rightarrow \left( \left| \overrightarrow{a}\times \overrightarrow{b} \right| \right)=\left| \overrightarrow{a} \right|\times \left| \overrightarrow{b} \right|\times \sin \theta \\\ & \Rightarrow \left( \overrightarrow{a}.\overrightarrow{b} \right)=\left| \overrightarrow{a} \right|\times \left| \overrightarrow{b} \right|\times \cos \theta \\\ \end{aligned}$$