Question

Let \[\overrightarrow{a}=\left( \overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\, \right)\] and \[\overrightarrow{b}=\left( \overset{\wedge }{\mathop{i}}\,+11\overset{\wedge }{\mathop{j}}\,+7\overset{\wedge }{\mathop{k}}\, \right)\] be given vectors. The vector \[\overrightarrow{r}=\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,+z\overset{\wedge }{\mathop{k}}\,\] that satisfies the equation \[\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\] is: -
(a) (1, -9, 14)
(b) (1, 9, 14)
(c) (1, 9, -14)
(d) (1, -9, -14)

Answer 1

Hint: Apply the cross – product of vectors r and a and equate it with b vector. Compare the coefficients of the unit vectors \[\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,\] and \[\overset{\wedge }{\mathop{k}}\,\] and form three linear equations in y and z. Solve any two equations to get the value of y and z. Solve any two equations to get the value of y and z. Use the formulas: - \[\overset{\wedge }{\mathop{i}}\,\times \overset{\wedge }{\mathop{i}}\,=\overset{\wedge }{\mathop{j}}\,\times \overset{\wedge }{\mathop{j}}\,=\overset{\wedge }{\mathop{k}}\,\times \overset{\wedge }{\mathop{k}}\,=0\], \[\overset{\wedge }{\mathop{i}}\,\times \overset{\wedge }{\mathop{j}}\,=\overset{\wedge }{\mathop{k}}\,\overset{\wedge }{\mathop{j}}\,\times \overset{\wedge }{\mathop{k}}\,=\overset{\wedge }{\mathop{i}}\,\] and \[\overset{\wedge }{\mathop{k}}\,\times \overset{\wedge }{\mathop{i}}\,=\overset{\wedge }{\mathop{j}}\,\].

Complete step by step answer:
Here, we have been provided with three vectors, \[\overrightarrow{a}=\overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,,\overrightarrow{b}=\overset{\wedge }{\mathop{i}}\,+11\overset{\wedge }{\mathop{j}}\,+7\overset{\wedge }{\mathop{k}}\,\] and \[\overrightarrow{r}=\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,+z\overset{\wedge }{\mathop{k}}\,\] which satisfies the relation \[\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\]. We have to find the values of y and z.
Now, let us consider the cross – product \[\overrightarrow{r}\times \overrightarrow{a}\]. So, we have,
\[\begin{align}
  & \Rightarrow \overrightarrow{r}\times \overrightarrow{a}=\left( \overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,+z\overset{\wedge }{\mathop{k}}\, \right)\times \left( \overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\, \right) \\
 & \Rightarrow \overrightarrow{r}\times \overrightarrow{a}=\left( \overset{\wedge }{\mathop{i}}\,\times \overset{\wedge }{\mathop{i}}\, \right)+\left( -2\times 1 \right)\left( \overset{\wedge }{\mathop{i}}\,\times \overset{\wedge }{\mathop{j}}\, \right)+\left( 1\times 3 \right)\left( \overset{\wedge }{\mathop{i}}\,\times \overset{\wedge }{\mathop{k}}\, \right)+\left( y\times 1 \right)\left( \overset{\wedge }{\mathop{j}}\,\times \overset{\wedge }{\mathop{i}}\, \right)+\left( y\times \left( -2 \right) \right)\left( \overset{\wedge }{\mathop{j}}\,\times \overset{\wedge }{\mathop{j}}\, \right) \\
 & +\left( y\times 3 \right)\left( \overset{\wedge }{\mathop{j}}\,\times \overset{\wedge }{\mathop{k}}\, \right)+\left( z\times 1 \right)\left( \overset{\wedge }{\mathop{k}}\,\times \overset{\wedge }{\mathop{i}}\, \right)+\left( z\times \left( -2 \right) \right)\left( \overset{\wedge }{\mathop{k}}\,\times \overset{\wedge }{\mathop{j}}\, \right)+\left( z\times 3 \right)\left( \overset{\wedge }{\mathop{k}}\,\times \overset{\wedge }{\mathop{k}}\, \right) \\
\end{align}\]
Now, we know that the cross – product of expression having same unit vectors is 0, so we have, \[\overset{\wedge }{\mathop{i}}\,\times \overset{\wedge }{\mathop{i}}\,=\overset{\wedge }{\mathop{j}}\,\times \overset{\wedge }{\mathop{j}}\,=\overset{\wedge }{\mathop{k}}\,\times \overset{\wedge }{\mathop{k}}\,=0\]. So, we have,
\[\Rightarrow \overrightarrow{r}\times \overrightarrow{a}=-2\left( \overrightarrow{i}\times \overrightarrow{j} \right)+3\left( \overrightarrow{i}\times \overrightarrow{k} \right)+\left( y\times 1 \right)\left( \overset{\wedge }{\mathop{j}}\,\times \overset{\wedge }{\mathop{i}}\, \right)+3y\left( \overrightarrow{j}\times \overrightarrow{k} \right)+z\left( \overrightarrow{k}\times \overrightarrow{i} \right)-2z\left( \overrightarrow{k}\times \overrightarrow{j} \right)\]
Now, applying the following listed cross – product formulas for the above relation, we have,
(i) \[\overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k},\overrightarrow{j}\times \overrightarrow{i}=-\overrightarrow{k}\]
(ii) \[\overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i},\overrightarrow{k}\times \overrightarrow{j}=-\overrightarrow{i}\]
(iii) \[\overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j},\overrightarrow{i}\times \overrightarrow{k}=-\overrightarrow{j}\]
\[\Rightarrow \overrightarrow{r}\times \overrightarrow{a}=-2\overrightarrow{k}-3\overrightarrow{j}-y\overrightarrow{k}+3y\overrightarrow{i}+z\overrightarrow{j}+2z\overrightarrow{i}\]
Grouping the terms of similar unit vectors, we get,
\[\Rightarrow \overrightarrow{r}\times \overrightarrow{a}=\left( 2z+3y \right)\overrightarrow{i}+\left( z-3 \right)\overrightarrow{j}+\left( -y-2 \right)\overrightarrow{k}\]
Now, applying the provided condition \[\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\], we get,
\[\begin{align}
  & \Rightarrow \overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b} \\
 & \Rightarrow \left( 2z+3y \right)\overrightarrow{i}+\left( z-3 \right)\overrightarrow{j}+\left( -y-2 \right)\overrightarrow{k}=\left( \overset{\wedge }{\mathop{i}}\,+11\overset{\wedge }{\mathop{j}}\,+7\overset{\wedge }{\mathop{k}}\, \right) \\
\end{align}\]
Comparing the coefficients of unit vectors \[\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,\] and \[\overset{\wedge }{\mathop{k}}\,\], we get,
\[\Rightarrow 2z+3y=1\] - (A)
\[\Rightarrow z-3=11\] - (B)
\[\Rightarrow -y-2=7\] - (C)
Now, considering equation (B), we get,
 \[\begin{align}
  & \Rightarrow z=11+3 \\
 & \Rightarrow z=14 \\
\end{align}\]
Considering equation (C), we get,
\[\begin{align}
  & \Rightarrow -y=7+2 \\
 & \Rightarrow -y=9 \\
 & \Rightarrow y=-9 \\
\end{align}\]
So, let us check if we have obtained the values of y and z as correct one or not. This can be done by substituting the values of y and z in equation (A). So, we have,
\[\begin{align}
  & \Rightarrow 2z+3y=2\times 14+3\times \left( -9 \right) \\
 & \Rightarrow 2z+3y=28-27 \\
 & \Rightarrow 2z+3y=1 \\
\end{align}\]
Hence, the values of y and z satisfies the equation (A). So, our obtained values are correct. Therefore, \[\overrightarrow{r}\] can be given as: -
\[\begin{align}
  & \Rightarrow \overrightarrow{r}=\overset{\wedge }{\mathop{i}}\,-9\overset{\wedge }{\mathop{j}}\,+14\overset{\wedge }{\mathop{k}}\, \\
 & \Rightarrow \left( y,z \right)=\left( -9,14 \right) \\
\end{align}\]

So, the correct answer is “Option a”.

Note: One may note that in vector \[\overrightarrow{r}\] the coefficient of \[\overset{\wedge }{\mathop{i}}\,\] is 1 which is already provided to us so we were required to find the values of variables y and z only. To find the values of y and z we have formed three equations (A), (B) and (C). Here, we have used equations (B) and (C) to find the values and equation (A) to check. You may use any two for finding the values and the third one to check. You must remember the listed cross – product of unit vectors in the given order otherwise you will make a sign mistake and the answer will be wrong. You may also use discriminant method to calculate the cross – product given as: - \[\overrightarrow{r}\times \overrightarrow{a}=\left| \begin{matrix}
   \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\
   1 & y & z \\
   1 & -2 & 3 \\
\end{matrix} \right|\]. The answer will be the same.