Question

Let \[\overrightarrow{a}=\widehat{i}-\widehat{j},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}\] and \[\overrightarrow{c}\] be a vector such that \[\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}=\overrightarrow{0}\] and \[\overrightarrow{a}.\overrightarrow{c}=4,\] then \[{{\left| \overrightarrow{c} \right|}^{2}}\] is equal to
\[\left( a \right)\dfrac{19}{2}\]
\[\left( b \right)8\]
\[\left( c \right)\dfrac{17}{2}\]
\[\left( d \right)9\]

Answer 1

Hint:We are given two vectors \[\overrightarrow{a}=\widehat{i}-\widehat{j},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}\] and we will use \[\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}=\overrightarrow{0}\] to get \[\overrightarrow{a}\times \overrightarrow{c}=-\overrightarrow{b}.\] Now, we will simplify further by applying the cross product on both the sides by \[\overrightarrow{a}.\] So, we will have \[\overrightarrow{a}\times \overrightarrow{c}\times \overrightarrow{a}=-\overrightarrow{b}\times \overrightarrow{a},\] then we will change \[-\overrightarrow{b}\times \overrightarrow{a}\] to \[\overrightarrow{a}\times \overrightarrow{b}.\] At last, we will open the triple product \[\overrightarrow{a}\times \overrightarrow{c}\times \overrightarrow{a}=\left( \overrightarrow{a}.\overrightarrow{a} \right).\overrightarrow{c}-\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{a}\] to find the vector c and \[{{\left| \overrightarrow{c} \right|}^{2}}.\]

We are given that we have two vectors \[\overrightarrow{a}=\widehat{i}-\widehat{j},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}.\] We have to find the vector c such that \[\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}=\overrightarrow{0}\] and \[\overrightarrow{a}.\overrightarrow{c}=4.\] Now, we are given that,
\[\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}=\overrightarrow{0}\]
So, we get,
\[\Rightarrow \overrightarrow{a}\times \overrightarrow{c}=-\overrightarrow{b}\]
Now, cross-product the above vector with vector a, we will get,
\[\left( \overrightarrow{a}\times \overrightarrow{c} \right)\times \overrightarrow{a}=-\overrightarrow{b}\times \overrightarrow{a}\]
For any vector X and Y, we know that,
\[\overrightarrow{X}\times \overrightarrow{Y}=-\overrightarrow{Y}\times \overrightarrow{X}\]
So,
\[-\overrightarrow{b}\times \overrightarrow{a}=\overrightarrow{a}\times \overrightarrow{b}\]
Hence, we have,
\[\left( \overrightarrow{a}\times \overrightarrow{c} \right)\times \overrightarrow{a}=\overrightarrow{a}\times \overrightarrow{b}.....\left( i \right)\]
Now, we have to find \[\overrightarrow{a}\times \overrightarrow{b}.\]
As we have, \[\overrightarrow{a}=\widehat{i}-\widehat{j},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k},\] so,
\[\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix}
   i & j & k \\
   1 & -1 & 0 \\
   1 & 1 & 1 \\
\end{matrix} \right|\]
Expanding along row 1, we will get,
\[\overrightarrow{a}\times \overrightarrow{b}=i\left( -1\times 1-0 \right)-j\left( 1\times 1-0 \right)+k\left( 1\times 1-\left( -1\times 1 \right) \right)\]
Solving further, we get,
\[\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=-i-j+2k\]
Using this in equation (i), we will get,
\[\left( \overrightarrow{a}\times \overrightarrow{c} \right)\times \overrightarrow{a}=-i-j+2k\]
Now, we know that our triple product is given as,
\[\left( \overrightarrow{A}\times \overrightarrow{B} \right)\times \overrightarrow{C}=\left( \overrightarrow{A}.\overrightarrow{C} \right).\overrightarrow{B}-\left( \overrightarrow{A}.\overrightarrow{B} \right).\overrightarrow{C}\]
So, \[\left( \overrightarrow{a}\times \overrightarrow{c} \right)\times \overrightarrow{a}\] is given as
\[\left( \overrightarrow{a}\times \overrightarrow{c} \right)\times \overrightarrow{a}=\left( \overrightarrow{a}.\overrightarrow{a} \right).\overrightarrow{c}-\left( \overrightarrow{c}.\overrightarrow{a} \right)\overrightarrow{a}\]
So, using this in \[\left( \overrightarrow{a}\times \overrightarrow{c} \right)\times \overrightarrow{a}=-i-j+2k\] we get,
\[\left( \overrightarrow{a}.\overrightarrow{a} \right).\overrightarrow{c}-\left( \overrightarrow{c}.\overrightarrow{a} \right).\overrightarrow{a}=-i-j+2k\]
\[\Rightarrow \left( \overrightarrow{a}.\overrightarrow{a} \right)=\left( i-j \right).\left( i-j \right)\]
\[\Rightarrow \left( \overrightarrow{a}.\overrightarrow{a} \right)=1+1\]
\[\Rightarrow \left( \overrightarrow{a}.\overrightarrow{a} \right)=2\]
And, \[\overrightarrow{c}.\overrightarrow{a}=4.\] So, we will get,
\[2\overrightarrow{c}-4\overrightarrow{a}=-i-j+2k\]
As, \[\overrightarrow{a}=\widehat{i}-\widehat{j}\] we will get,
\[2\overrightarrow{c}=-i-j+2k+4\left( i-j \right)\]
Simplifying, we get,
\[2\overrightarrow{c}=3i-5j+2k\]
Dividing both the sides by 2, we will get,
\[\Rightarrow \overrightarrow{c}=\dfrac{3}{2}i-\dfrac{5}{2}j+k\]
Now,
\[{{\left| \overrightarrow{c} \right|}^{2}}=\overrightarrow{c}.\overrightarrow{c}\]
\[\Rightarrow {{\left| \overrightarrow{c} \right|}^{2}}=\left( \dfrac{3}{2}i-\dfrac{5}{2}j+k \right)\left( \dfrac{3}{2}i-\dfrac{5}{2}j+k \right)\]
After simplification, we will get,
\[\Rightarrow {{\left| \overrightarrow{c} \right|}^{2}}=\dfrac{3}{2}\times \dfrac{3}{2}+\left( \dfrac{-5}{2}\times \dfrac{-5}{2} \right)+1\times 1\]
\[\Rightarrow {{\left| \overrightarrow{c} \right|}^{2}}=\dfrac{9}{4}+\dfrac{25}{4}+1\]
Solving further, we get,
\[\Rightarrow {{\left| \overrightarrow{c} \right|}^{2}}=\dfrac{38}{4}\]
\[\Rightarrow {{\left| \overrightarrow{c} \right|}^{2}}=\dfrac{19}{2}\]
Hence, option (a) is the right answer.

Note: The dot product of two vectors is defined as the product of the sum of the product of the corresponding vector entries. If a = xi + yj and b = ci + dj, we get,
\[a.b=\left( xi+yj \right)\left( ci+dj \right)\]
\[\Rightarrow a.b=xc+yd\]
That’s, why,
\[\overrightarrow{c}.\overrightarrow{c}=\left( \dfrac{3}{2}i-\dfrac{5}{2}j+k \right).\left( \dfrac{3}{2}i-\dfrac{5}{2}j+k \right)\]
We get,
\[\Rightarrow \overrightarrow{c}.\overrightarrow{c}=\dfrac{3}{2}\times \dfrac{3}{2}+\left( \dfrac{-5}{2}\times \dfrac{-5}{2} \right)+1\times 1\]
\[\Rightarrow \overrightarrow{c}.\overrightarrow{c}=\dfrac{9}{4}+\dfrac{25}{4}+1\]
\[\Rightarrow \overrightarrow{c}.\overrightarrow{c}=\dfrac{9+25+4}{4}\]
\[\Rightarrow \overrightarrow{c}.\overrightarrow{c}=\dfrac{38}{4}\]
Simplifying, we get,
\[\Rightarrow {{\left| \overrightarrow{c} \right|}^{2}}=\dfrac{19}{2}\]