Question

Let $\overrightarrow{a}=3\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\,$ and $\overrightarrow{b}=\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\,$ be two vectors. If a vector perpendicular to both the vectors $\overrightarrow{a}+\overrightarrow{b}\And \overrightarrow{a}-\overrightarrow{b}$ has the magnitude 12 then one such vector is:
(a) $4\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right)$
(b) $4\left( 2\overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right)$
(c) $4\left( 2\overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)$
(d) $4\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)$

Answer 1

Hint: We have given two vectors a and b then find the vectors $\overrightarrow{a}+\overrightarrow{b}\And \overrightarrow{a}-\overrightarrow{b}$ by adding and subtracting the two vectors respectively. The addition and subtraction of two vectors is done by adding and subtracting the corresponding x, y and z unit vectors. Now, we have to find the vector which is perpendicular to both $\overrightarrow{a}+\overrightarrow{b}\And \overrightarrow{a}-\overrightarrow{b}$ which we are going to find by taking cross product of these two vectors which will give you the direction of the vector and then multiply by 12 to get the complete vector.

Complete step-by-step answer:
We have given below two vectors as follows:
$\overrightarrow{a}=3\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\,$
$\overrightarrow{b}=\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\,$
Now, adding the above two vectors by adding their corresponding units vectors lying in x, y and z direction we get,
$\begin{align}
  & \overrightarrow{a}+\overrightarrow{b}=\left( 3+1 \right)\overset{\wedge }{\mathop{i}}\,+\left( 2+2 \right)\overset{\wedge }{\mathop{j}}\,+\left( 2-2 \right)\overset{\wedge }{\mathop{k}}\, \\
 & \Rightarrow \overrightarrow{a}+\overrightarrow{b}=4\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\, \\
\end{align}$
Now, subtracting the two vectors a and b we get,
$\begin{align}
  & \overrightarrow{a}-\overrightarrow{b}=\left( 3-1 \right)\overset{\wedge }{\mathop{i}}\,+\left( 2-2 \right)\overset{\wedge }{\mathop{j}}\,+\left( 2-\left( -2 \right) \right)\overset{\wedge }{\mathop{k}}\, \\
 & \Rightarrow \overrightarrow{a}-\overrightarrow{b}=2\overset{\wedge }{\mathop{i}}\,+0\overset{\wedge }{\mathop{j}}\,+4\overset{\wedge }{\mathop{k}}\, \\
 & \Rightarrow \overrightarrow{a}-\overrightarrow{b}=2\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{k}}\, \\
\end{align}$
Now, taking the cross product of $\overrightarrow{a}+\overrightarrow{b}\And \overrightarrow{a}-\overrightarrow{b}$ we get,
Let us assume we have two vectors $\overrightarrow{x}={{x}_{1}}\overset{\wedge }{\mathop{i}}\,+{{x}_{2}}\overset{\wedge }{\mathop{j}}\,+{{x}_{3}}\overset{\wedge }{\mathop{k}}\,$ and $\overrightarrow{y}={{y}_{1}}\overset{\wedge }{\mathop{i}}\,+{{y}_{2}}\overset{\wedge }{\mathop{j}}\,+{{y}_{3}}\overset{\wedge }{\mathop{k}}\,$ then the cross product of the two vectors is given as:
$\left( \overrightarrow{x} \right)\times \left( \overrightarrow{y} \right)=\left| \begin{matrix}
   \overset{\wedge }{\mathop{i}}\, & \overset{\wedge }{\mathop{j}}\, & \overset{\wedge }{\mathop{k}}\, \\
   {{x}_{1}} & {{x}_{2}} & {{x}_{3}} \\
   {{y}_{1}} & {{y}_{2}} & {{y}_{3}} \\
\end{matrix} \right|$
Then using the above relation we can find the cross product of $\overrightarrow{a}+\overrightarrow{b}\And \overrightarrow{a}-\overrightarrow{b}$.
$\left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{a}-\overrightarrow{b} \right)=\left| \begin{matrix}
   \overset{\wedge }{\mathop{i}}\, & \overset{\wedge }{\mathop{j}}\, & \overset{\wedge }{\mathop{k}}\, \\
   4 & 4 & 0 \\
   2 & 0 & 4 \\
\end{matrix} \right|$
Now, expanding the determinant written on the right hand side of the above equation we get,
\[\begin{align}
  & \left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{a}-\overrightarrow{b} \right)=\overset{\wedge }{\mathop{i}}\,\left( 4\left( 4 \right)-0\left( 0 \right) \right)-\overset{\wedge }{\mathop{j}}\,\left( 4\left( 4 \right)-2\left( 0 \right) \right)+\overset{\wedge }{\mathop{k}}\,\left( 4\left( 0 \right)-4\left( 2 \right) \right) \\
 & \Rightarrow \left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{a}-\overrightarrow{b} \right)=\overset{\wedge }{\mathop{i}}\,\left( 16 \right)-\overset{\wedge }{\mathop{j}}\,\left( 16 \right)+\overset{\wedge }{\mathop{k}}\,\left( -8 \right) \\
\end{align}\]
Taking 8 as common the right hand side we get,
\[\left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{a}-\overrightarrow{b} \right)=8\left( \overset{\wedge }{\mathop{i}}\,\left( 2 \right)-\overset{\wedge }{\mathop{j}}\,\left( 2 \right)+\overset{\wedge }{\mathop{k}}\,\left( -1 \right) \right)\]
Magnitude of the above \[\left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{a}-\overrightarrow{b} \right)\] vector is equal to:
$\begin{align}
  & \sqrt{{{8}^{2}}\left( {{2}^{2}}+{{2}^{2}}+1 \right)} \\
 & =8\sqrt{4+4+1} \\
 & =8\sqrt{9} \\
 & =8\left( 3 \right) \\
 & =24 \\
\end{align}$
Now, we have to find the unit vector of the above vector by dividing the magnitude of this vector \[\left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{a}-\overrightarrow{b} \right)\].
\[\begin{align}
  & \dfrac{8\left( \overset{\wedge }{\mathop{i}}\,\left( 2 \right)-\overset{\wedge }{\mathop{j}}\,\left( 2 \right)+\overset{\wedge }{\mathop{k}}\,\left( -1 \right) \right)}{24} \\
 & =\dfrac{1}{3}\left( \overset{\wedge }{\mathop{i}}\,\left( 2 \right)-\overset{\wedge }{\mathop{j}}\,\left( 2 \right)+\overset{\wedge }{\mathop{k}}\,\left( -1 \right) \right) \\
\end{align}\]
Now, we have direction of the required vector and we have also given the magnitude of the required vector as 12 so we know that a vector is the multiplication of direction and magnitude so we are going to multiply the above unit vector with 12 we get,
\[\begin{align}
  & \dfrac{1}{3}\left( \overset{\wedge }{\mathop{i}}\,\left( 2 \right)-\overset{\wedge }{\mathop{j}}\,\left( 2 \right)+\overset{\wedge }{\mathop{k}}\,\left( -1 \right) \right)\left( 12 \right) \\
 & =4\left( \overset{\wedge }{\mathop{2i}}\,-2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right) \\
\end{align}\]
Hence, the correct option is (b).

Note: The mistake that you could make in this question is that you might think that the vector that question is demanding is the vector that will come from the cross product of $\overrightarrow{a}+\overrightarrow{b}\And \overrightarrow{a}-\overrightarrow{b}$ but actually, the cross product is only giving the direction of the vector which we have calculated by the unit vector along that direction so make sure you won’t make this mistake in the question.