Question

If u = a – b and v = a + b and |a| = |b| = 2, then $\left| u\times v \right|$ is equal to (Note: a and b are vectors).
$\left( a \right)2\sqrt{16-{{\left( a.b \right)}^{2}}}$
$\left( b \right)\sqrt{16-{{\left( a.b \right)}^{2}}}$
$\left( c \right)2\sqrt{4-{{\left( a.b \right)}^{2}}}$
$\left( d \right)2\sqrt{4+{{\left( a.b \right)}^{2}}}$

Answer 1

To solve this we will first try to calculate the value of $\left| u\times v \right|$ in terms of a and b. For that we will substitute u = a – b and v = a + b and solve using the formula, ${{\left| a\times b \right|}^{2}}+{{\left( a.b \right)}^{2}}={{\left( ab\sin \theta \right)}^{2}}+{{\left( ab\cos \theta \right)}^{2}}$ and ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.$ Finally, we will use $\left| a \right|=\left| b \right|=2$ to get the result.

Complete step-by-step answer:
We are given that, u = a – b and v = a + b and |a| = 2 and |b| = 2. Then to find the value of $\left| u\times v \right|$ we have u = a – b and v = a + b.
$\Rightarrow \left| u\times v \right|=\left| \left( a-b \right)\times \left( a+b \right) \right|$
Now, any vector cross product with itself is 0.
$\Rightarrow a\times a=0;b\times b=0$
$\Rightarrow \left| u\times v \right|=2\left| a\times b \right|.....\left( i \right)$
Now, we have a formula relating ${{\left| a\times b \right|}^{2}}$ and ${{\left( a.b \right)}^{2}}$ with $\sin \theta$ and $\cos \theta .$ It is given as,
${{\left| a\times b \right|}^{2}}+{{\left( a.b \right)}^{2}}={{\left( ab\sin \theta \right)}^{2}}+{{\left( ab\cos \theta \right)}^{2}}$
$\Rightarrow {{\left| a\times b \right|}^{2}}+{{\left( a.b \right)}^{2}}={{a}^{2}}{{b}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)$
As, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.$
$\Rightarrow {{\left| a\times b \right|}^{2}}+{{\left( a.b \right)}^{2}}={{a}^{2}}{{b}^{2}}\times 1$
$\Rightarrow {{\left| a\times b \right|}^{2}}+{{\left( a.b \right)}^{2}}={{a}^{2}}{{b}^{2}}$
$\Rightarrow {{\left| a\times b \right|}^{2}}={{a}^{2}}{{b}^{2}}-{{\left( a.b \right)}^{2}}$
$\Rightarrow {{\left| a\times b \right|}^{2}}=\sqrt{{{a}^{2}}{{b}^{2}}-{{\left( a.b \right)}^{2}}}.....\left( ii \right)$
Now, let us substitute the value obtained in equation (i) in equation (ii).
$\Rightarrow \left| u\times v \right|=2\left| a\times b \right|$
$\Rightarrow \left| u\times v \right|=2\sqrt{{{a}^{2}}{{b}^{2}}-{{\left( ab \right)}^{2}}}$
Now, as $\left| a \right|=\left| b \right|=2,$
${{a}^{2}}={{2}^{2}}$
$\left| {{b}^{2}} \right|={{\left| b \right|}^{2}}={{b}^{2}}$
${{b}^{2}}={{2}^{2}}$
Substituting these values in the above equation, we get,
$\Rightarrow \left| u\times v \right|=2\sqrt{{{2}^{2}}{{2}^{2}}-{{\left( a.b \right)}^{2}}}$
Hence, the value of $\left| u\times v \right|=2\sqrt{16-{{\left( a.b \right)}^{2}}}.$

So, the correct answer is “Option a”.

Note: The point of confusion can be using (a) = 2 in place of |a| = 2. This is correct as ${{\left| a \right|}^{2}}={{a}^{2}}$ as ‘a’ be any value positive or negative, ${{a}^{2}}$ always will be positive $\Rightarrow {{\left| a \right|}^{2}}={{a}^{2}}$ as $\left| a \right|\ge 0,$ ‘a’ be any value. The point is we cannot use a = 2 directly. We need to use ${{a}^{2}}={{2}^{2}}$ but not a = 2, as |a| = 2, ‘a’ can be negative 2 or positive as well.