Question

Let \[a = 2i + j - 2k\]and \[b = i + j\]. If c is a vector such that \[a.c = |c|\], \[|c - a| = 2\sqrt 2 \]and the angle between \[a \times b\] and c is \[{30^0}\], then \[|(a \times b) \times c|\] is equal to?
A. \[\dfrac{2}{3}\]
B. \[\dfrac{3}{2}\]
C. 2
D. 3

Answer 1

Hint: Here in this question, we have to find the value of \[|(a\times b)\times c|\], so initially we have to find the value of \[a\times b\] and then \[c\], then we can find the cross product of the both. To find the value of the cross product of two vectors we use the concepts of determinants.

Formula Used:
 \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]

Complete step by step explanation:
Let us consider the given data, \[a=2i+j-2k\] and \[b=i+j\]
We have to find the value of \[|(a\times b)\times c|\]
\[\Rightarrow |(a\times b)\times c|=|a\times b||c|\sin \theta \]
The angle between \[a\times b\] and c is \[{{30}^{0}}\], so we have
\[\Rightarrow \,|(a\times b)\times c|\,=\,|a\times b||c|\sin {{30}^{0}}\]………………….. (1)
Now we will find the value of \[|a\times b|\]
\[a\times b=\left| \begin{matrix}
   i & j & k \\
   2 & 1 & -2 \\
   1 & 1 & 0 \\
\end{matrix} \right|\]
\[=i(0+2)-j(0+2)+k(2-1)\]
\[=2i-2j+k\]
\[|a\times b|=\sqrt{{{2}^{2}}+{{2}^{2}}+{{1}^{2}}}\]
\[\Rightarrow |a\times b|=\sqrt{4+4+1}\]
\[\Rightarrow |a\times b|=\sqrt{9}\]
\[\Rightarrow |a\times b|=3\]……………… (2)
Now consider \[|c-a|=2\sqrt{2}\]
Squaring on both sides
\[\Rightarrow |c-a{{|}^{2}}={{\left( 2\sqrt{2} \right)}^{2}}\]
Apply the formula of \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]
\[\Rightarrow |c{{|}^{2}}+|a{{|}^{2}}-2ac=8\]
As given \[a.c=|c|\] and \[|a{{|}^{2}}=4+4+1=9\], so we have
\[\Rightarrow |c{{|}^{2}}+9-2|c|=8\]
simplify
\[\Rightarrow |c{{|}^{2}}-2|c|+1=0\]
Solve the equation
\[\Rightarrow {{\left( |c|-1 \right)}^{2}}=0\]
\[\Rightarrow |c|=1\]……………. (3)
 substitute (2) and (3) in equation (1) we get
\[\Rightarrow \,|(a\times b)\times c|\,=\,3.1\sin {{30}^{0}}\]
As we know that the value of \[\sin {{30}^{0}}=\dfrac{1}{2}\] and now we have
\[\Rightarrow \,|(a\times b)\times c|\,=\,3.1.\dfrac{1}{2}\]
\[\Rightarrow \,|(a\times b)\times c|\,=\,\dfrac{3}{2}\]
Therefore the value of \[\,|(a\times b)\times c|\,\] is \[\dfrac{3}{2}\]

Hence the option B is correct one.

Additional Information: An item that has both magnitude and direction is referred to as a vector. It is often represented by an arrow whose length is proportional to the magnitude of the quantity and whose direction is the same as that of the quantity. A vector does not have a position while having magnitude and direction.

Note: If the magnitude and direction of any two vectors are the same, they can be said to be identical vectors. The force exerted on an object serves as the best illustration of a vector since both the magnitude and the direction of the applied force have an impact on the object's response. It is impossible to alter a vector's magnitude by rotating or moving it around.