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

Hint: 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.