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Question: Let \[\overrightarrow{a}=\widehat{i}-\widehat{j},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat...

Let a→=i^−j^,b→=i^+j^+k^\overrightarrow{a}=\widehat{i}-\widehat{j},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k} and c→\overrightarrow{c} be a vector such that a→×c→+b→=0→\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}=\overrightarrow{0} and a→.c→=4,\overrightarrow{a}.\overrightarrow{c}=4, then ∣c→∣2{{\left| \overrightarrow{c} \right|}^{2}} is equal to
(a)192\left( a \right)\dfrac{19}{2}
(b)8\left( b \right)8
(c)172\left( c \right)\dfrac{17}{2}
(d)9\left( d \right)9

Explanation

Solution

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

We are given that we have two vectors a→=i^−j^,b→=i^+j^+k^.\overrightarrow{a}=\widehat{i}-\widehat{j},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}. We have to find the vector c such that a→×c→+b→=0→\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}=\overrightarrow{0} and a→.c→=4.\overrightarrow{a}.\overrightarrow{c}=4. Now, we are given that,
a→×c→+b→=0→\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}=\overrightarrow{0}
So, we get,
⇒a→×c→=−b→\Rightarrow \overrightarrow{a}\times \overrightarrow{c}=-\overrightarrow{b}
Now, cross-product the above vector with vector a, we will get,
(a→×c→)×a→=−b→×a→\left( \overrightarrow{a}\times \overrightarrow{c} \right)\times \overrightarrow{a}=-\overrightarrow{b}\times \overrightarrow{a}
For any vector X and Y, we know that,
X→×Y→=−Y→×X→\overrightarrow{X}\times \overrightarrow{Y}=-\overrightarrow{Y}\times \overrightarrow{X}
So,
−b→×a→=a→×b→-\overrightarrow{b}\times \overrightarrow{a}=\overrightarrow{a}\times \overrightarrow{b}
Hence, we have,
(a→×c→)×a→=a→×b→.....(i)\left( \overrightarrow{a}\times \overrightarrow{c} \right)\times \overrightarrow{a}=\overrightarrow{a}\times \overrightarrow{b}.....\left( i \right)
Now, we have to find a→×b→.\overrightarrow{a}\times \overrightarrow{b}.
As we have, a→=i^−j^,b→=i^+j^+k^,\overrightarrow{a}=\widehat{i}-\widehat{j},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}, so,

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}$$