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

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:
$\begin{aligned} & \overset{\to }{\mathop{r}}\,=\left( \hat{i}+2\hat{j}-\hat{k} \right)+\lambda \left( \hat{i}+\hat{j}-2\hat{k} \right) \\\ & =\left( 1+\lambda \right)\hat{i}+\left( 2+\lambda \right)\hat{j}+\left( -1-2\lambda \right)\hat{k}......(2) \end{aligned}$
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:

& \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{aligned}$$ Now, we need to find the magnitude of $\overset{\to }{\mathop{a}}\,$ We know that: $$\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: $\begin{aligned} & \left| \overset{\to }{\mathop{a}}\, \right|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\\ & =\sqrt{4+1+1} \\\ & =\sqrt{6}......(4) \end{aligned}$ 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): $\begin{aligned} & \dfrac{\left( -1-\lambda \right)}{\sqrt{6}}=\sqrt{\dfrac{2}{3}} \\\ & \left( -1-\lambda \right)=\pm 2 \\\ & -\lambda =3,-1 \\\ & \lambda =-3,1 \\\ \end{aligned}$ Now, put the value of $\lambda $in equation (2), we get: For $\lambda =-3$ $\begin{aligned} & \overset{\to }{\mathop{r}}\,=\left( 1-3 \right)\hat{i}+\left( 2-3 \right)\hat{j}+\left( -1+6 \right)\hat{k} \\\ & =-2\hat{i}-\hat{j}+5\hat{k} \end{aligned}$ For $\lambda =1$ $\begin{aligned} & \overset{\to }{\mathop{r}}\,=\left( 1+1 \right)\hat{i}+\left( 2+1 \right)\hat{j}+\left( -1-2 \right)\hat{k} \\\ & =2\hat{i}+3\hat{j}-3\hat{k} \end{aligned}$ **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: $pro{{j}_{b}}a=\dfrac{\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,}{\overset{\to }{\mathop{a}}\,}$