Question

Let $\overset{\to }{\mathop{a}}\,=2\hat{i}-\hat{j}+\hat{k};\overset{\to }{\mathop{b}}\,=\hat{i}+2\hat{j}-\hat{k}$and $\overset{\to }{\mathop{c}}\,=\hat{i}+\hat{j}-2\hat{k}$ be three vectors. A vector in the plane of b and c whose projection on a is of magnitude $\sqrt{\dfrac{2}{3}}$ is
a)$2\hat{i}+2\hat{j}-3\hat{k}$
b) $2\hat{i}+3\hat{j}+3\hat{k}$
c) $2\hat{i}-\hat{j}+5\hat{k}$
d) $2\hat{i}+\hat{j}+5\hat{k}$

Answer 1

Hint: We have three vectors given as: $\overset{\to }{\mathop{a}}\,=2\hat{i}-\hat{j}+\hat{k};\overset{\to }{\mathop{b}}\,=\hat{i}+2\hat{j}-\hat{k}$and $\overset{\to }{\mathop{c}}\,=\hat{i}+\hat{j}-2\hat{k}$. Let us assume that $\overset{\to }{\mathop{r}}\,=\overset{\to }{\mathop{b}}\,+\lambda \overset{\to }{\mathop{c}}\,$ is the vector in the plane of b and c. Find the vector r and then find the projection of $\overset{\to }{\mathop{r}}\,$on $\overset{\to }{\mathop{a}}\,$ by using the projection formula as: $\dfrac{\overset{\to }{\mathop{r}}\,.\overset{\to }{\mathop{a}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|}$. We know that the projection of $\overset{\to }{\mathop{r}}\,$on $\overset{\to }{\mathop{a}}\,$ is $\sqrt{\dfrac{2}{3}}$. So, find the value of $\lambda $ using this relation and then find the required vector $\overset{\to }{\mathop{r}}\,$.

Complete step-by-step solution:
As we have assumed that: $\overset{\to }{\mathop{r}}\,=\overset{\to }{\mathop{b}}\,+\lambda \overset{\to }{\mathop{c}}\,......(1)$
Now, put value of $\overset{\to }{\mathop{b}}\,=\hat{i}+2\hat{j}-\hat{k}$ and $\overset{\to }{\mathop{c}}\,=\hat{i}+\hat{j}-2\hat{k}$ in equation (1), we get:
$\begin{align}
  & \overset{\to }{\mathop{r}}\,=\left( \hat{i}+2\hat{j}-\hat{k} \right)+\lambda \left( \hat{i}+\hat{j}-2\hat{k} \right) \\
 & =\left( 1+\lambda \right)\hat{i}+\left( 2+\lambda \right)\hat{j}+\left( -1-2\lambda \right)\hat{k}......(2)
\end{align}$
Now, we need to find the projection of $\overset{\to }{\mathop{r}}\,$on $\overset{\to }{\mathop{a}}\,$ by using the formula: $\dfrac{\overset{\to }{\mathop{r}}\,.\overset{\to }{\mathop{a}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|}$
Firstly, find $\overset{\to }{\mathop{r}}\,.\overset{\to }{\mathop{a}}\,$, we get:
\[\begin{align}
  & \overset{\to }{\mathop{r}}\,.\overset{\to }{\mathop{a}}\,=\left\{ \left( 1+\lambda \right)\hat{i}+\left( 2+\lambda \right)\hat{j}+\left( -1-2\lambda \right)\hat{k} \right\}.\left\{ 2\hat{i}-\hat{j}+\hat{k} \right\} \\
 & =2\left( 1+\lambda \right)-1\left( 2+\lambda \right)+1\left( -1-2\lambda \right) \\
 & =2+2\lambda -2-\lambda -1-2\lambda \\
 & =-1-\lambda ......(3)
\end{align}\]
Now, we need to find the magnitude of $\overset{\to }{\mathop{a}}\,$
We know that: \[\left| \overset{\to }{\mathop{a}}\, \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\]
So, for $\overset{\to }{\mathop{a}}\,=2\hat{i}-\hat{j}+\hat{k}$ we get:
$\begin{align}
  & \left| \overset{\to }{\mathop{a}}\, \right|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\
 & =\sqrt{4+1+1} \\
 & =\sqrt{6}......(4)
\end{align}$
Now, by using the projection formula: $\dfrac{\overset{\to }{\mathop{r}}\,.\overset{\to }{\mathop{a}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|}$, we get:
$\dfrac{\overset{\to }{\mathop{r}}\,.\overset{\to }{\mathop{a}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|}=\dfrac{\left( -1-\lambda \right)}{\sqrt{6}}......(5)$
We know that the projection of $\overset{\to }{\mathop{r}}\,$on $\overset{\to }{\mathop{a}}\,$ is $\sqrt{\dfrac{2}{3}}$.
So, we can wite equation (5):
$\begin{align}
  & \dfrac{\left( -1-\lambda \right)}{\sqrt{6}}=\sqrt{\dfrac{2}{3}} \\
 & \left( -1-\lambda \right)=\pm 2 \\
 & -\lambda =3,-1 \\
 & \lambda =-3,1 \\
\end{align}$
Now, put the value of $\lambda $in equation (2), we get:
For $\lambda =-3$
$\begin{align}
  & \overset{\to }{\mathop{r}}\,=\left( 1-3 \right)\hat{i}+\left( 2-3 \right)\hat{j}+\left( -1+6 \right)\hat{k} \\
 & =-2\hat{i}-\hat{j}+5\hat{k}
\end{align}$
For $\lambda =1$
$\begin{align}
  & \overset{\to }{\mathop{r}}\,=\left( 1+1 \right)\hat{i}+\left( 2+1 \right)\hat{j}+\left( -1-2 \right)\hat{k} \\
 & =2\hat{i}+3\hat{j}-3\hat{k}
\end{align}$
Hence, option (a) is the correct answer.

Note: The vector projection is of two types: Scalar projection that tells about the magnitude of the vector projection and the other is the Vector projection which says about itself and represents the unit vector.
If the vector $\overset{\to }{\mathop{a}}\,$ is projected on vector $\overset{\to }{\mathop{b}}\,$ then Vector Projection formula is given below:
$pro{{j}_{b}}a=\dfrac{\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,}{{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}}\overset{\to }{\mathop{b}}\,$
The Scalar projection formula defines the length of given vector projection and is given below:
$pro{{j}_{b}}a=\dfrac{\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,}{\overset{\to }{\mathop{a}}\,}$