Question

Let $\vec{a}=2\hat{i}+\hat{j}-\hat{k}\text{ and }\vec{b}=\hat{i}+2\hat{j}+\hat{k}$ be two vectors. Consider a vector $\vec{c}=\alpha \vec{a}+\beta \vec{b},\alpha ,\beta \in R$. If the projection of $\vec{c}$ on vector $\left( \vec{a}+\vec{b} \right)$ is $3\sqrt{2}$ then minimum value of $\left( \vec{c}-\left( \vec{a}\times \vec{b} \right) \right)\cdot \vec{c}$ is equal to ...

Answer 1

Hint: In this question, we are given three vectors $\vec{a},\vec{b},\vec{c}$. We have to find minimum value $\left( \vec{c}-\left( \vec{a}\times \vec{b} \right) \right)\cdot \vec{c}$ when projection of $\vec{c}$ on vector $\left( \vec{a}+\vec{b} \right)$ is given. For this, we will first calculate projection of $\vec{c}$ on $\left( \vec{a}+\vec{b} \right)$ in our own terms to find value of a+b which will be used in $\left( \vec{c}-\left( \vec{a}\times \vec{b} \right) \right)\cdot \vec{c}$. We will use following properties of vectors to evaluate our answer:
(I) Projection of $\vec{B}$ on $\vec{A}$ is given as $\dfrac{{\vec{A}}}{\left| {\vec{A}} \right|}\cdot \vec{B}$ where $\left| {\vec{A}} \right|$ represents the magnitude of $\vec{A}$.
(II) Magnitude of $x\hat{i}+y\hat{j}+z\hat{k}$ is given as $\left| {\vec{A}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$.
(III) Dot product of $\vec{A}={{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}+{{z}_{1}}\hat{k}\text{ and }\vec{B}={{x}_{2}}\hat{i}+{{y}_{2}}\hat{j}+{{z}_{2}}\hat{k}$ is given as $\vec{A}\cdot \vec{B}={{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+{{z}_{1}}{{z}_{2}}$.
(IV) $\vec{a}\cdot \vec{a}={{\left| {\vec{a}} \right|}^{2}}$ where $\left| {\vec{a}} \right|$ is the magnitude of $\vec{a}$.
(V) $\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$ which is commutative property.
(VI) $\left( \vec{a}\times \vec{b} \right)\cdot \vec{c}$ is equal to zero if any two of the vectors are equal.

Complete step-by-step solution:
Here, we are given vector $\vec{a}$ as $2\hat{i}+\hat{j}-\hat{k}$ and vector $\vec{b}$ as $\hat{i}+2\hat{j}+\hat{k}$. Vector $\vec{c}$ is given as $\vec{c}=\alpha \vec{a}+\beta \vec{b}$ where $\alpha ,\beta \in R$.
We know, projection of any vector $\vec{B}$ on \[\vec{A}\] is given by $\dfrac{{\vec{A}}}{\left| {\vec{A}} \right|}\cdot \vec{B}$.
Here, projection of $\vec{c}$ on $\left( \vec{a}+\vec{b} \right)$ is given as $3\sqrt{2}$ therefore,
\[\Rightarrow \dfrac{\left( \vec{a}+\vec{b} \right)}{\left( \left| \vec{a}+\vec{b} \right| \right)}\cdot \vec{c}=3\sqrt{2}\cdots \cdots \cdots \cdots \left( 1 \right)\]
Let us evaluate left side,
$\left( \vec{a}+\vec{b} \right)$ will be equal to $\Rightarrow 2\hat{i}+\hat{j}-\hat{k}+\hat{i}+2\hat{j}+\hat{k}\Rightarrow 3\hat{i}+3\hat{j}$.
$\left| \vec{a}+\vec{b} \right|$ represent magnitude of $\left( \vec{a}+\vec{b} \right)$ and magnitude of any vector \[\vec{A}=x\hat{i}+y\hat{j}+z\hat{k}\] is given by $\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$ therefore, $\Rightarrow \left| \vec{a}+\vec{b} \right|=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 3 \right)}^{2}}}=\sqrt{9+9}=\sqrt{18}$.
Now, $\sqrt{18}$ can be written as $\Rightarrow \sqrt{3\times 3\times 2}=3\sqrt{2}$.
So, $\Rightarrow \left| \vec{a}+\vec{b} \right|=3\sqrt{2}$.
Now, $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$ so $\vec{a}\cdot \vec{a}=\left( 2\hat{i}+\hat{j}-\hat{k} \right)\cdot \left( 2\hat{i}+\hat{j}-\hat{k} \right)$.
We know $\vec{a}\cdot \vec{a}={{\left| {\vec{a}} \right|}^{2}}$,$\therefore {{\left| {\vec{a}} \right|}^{2}}=4+1+1=6$.
Therefore, $\Rightarrow {{\left| {\vec{a}} \right|}^{2}}=6$.
And $\vec{b}=\hat{i}+2\hat{j}+\hat{k}$ so
 $\begin{align}
&\Rightarrow \vec{b}\cdot \vec{b}=\left( \hat{i}+2\hat{j}+\hat{k} \right)\cdot \left( \hat{i}+2\hat{j}+\hat{k} \right) \\
 & \Rightarrow {{\left| {\vec{b}} \right|}^{2}}=1+4+1=6 \\
\end{align}$
Therefore, $\Rightarrow {{\left| {\vec{b}} \right|}^{2}}=6$.
Now $\left( \vec{a}+\vec{b} \right)\cdot \vec{c}$ is equal to $\left( \vec{a}+\vec{b} \right)\cdot \left( \alpha \vec{a}+\beta \vec{b} \right)$.
\[\Rightarrow \alpha \vec{a}\cdot \vec{a}+\alpha \vec{b}\cdot \vec{a}+\beta \vec{a}\cdot \vec{b}+\beta \vec{b}\cdot \vec{b}\]
We know, $\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$ so we get:
\[\Rightarrow \alpha {{\left| {\vec{a}} \right|}^{2}}+\alpha \left( \vec{a}\cdot \vec{b} \right)+\beta \left( \vec{a}\cdot \vec{b} \right)+\beta {{\left| {\vec{b}} \right|}^{2}}\cdots \cdots \cdots \cdots \left( 2 \right)\]
$\vec{a}\cdot \vec{b}$ can be calculated as $\Rightarrow \left( 2\hat{i}+\hat{j}-\hat{k} \right)\cdot \left( \hat{i}+2\hat{j}+\hat{k} \right)=2+2-1=3$.
So, $\Rightarrow \vec{a}\cdot \vec{b}=3$.
Using value of ${{\left| {\vec{a}} \right|}^{2}},{{\left| {\vec{b}} \right|}^{2}}\text{ and }\left( \vec{a}\cdot \vec{b} \right)$ in (2) we get:
\[\begin{align}
  & \Rightarrow \alpha \left( 6 \right)+\alpha \left( 3 \right)+\beta \left( 3 \right)+\beta \left( 6 \right) \\
 & \Rightarrow 9\alpha +9\beta \\
 & \Rightarrow 9\left( \alpha +\beta \right) \\
\end{align}\]
Now, put value of $\left( \vec{a}+\vec{b} \right)\cdot \vec{c}\text{ and }\left| \vec{a}+\vec{b} \right|$ in (1) we get:
\[\begin{align}
  & \Rightarrow \dfrac{9\left( \alpha +\beta \right)}{3\sqrt{2}}=3\sqrt{2} \\
 & \Rightarrow 9\left( \alpha +\beta \right)=3\sqrt{2}\times 3\sqrt{2} \\
 & \Rightarrow 9\left( \alpha +\beta \right)=18 \\
 & \Rightarrow \alpha +\beta =2 \\
\end{align}\]
Let us evaluate $\left( \vec{c}-\left( \vec{a}\times \vec{b} \right) \right)\cdot \vec{c}$. Putting the value of $\vec{c}$.
\[\begin{align}
  & \Rightarrow \left( \alpha \vec{a}+\beta \vec{b}-\left( \vec{a}\times \vec{b} \right) \right)\cdot \left( \alpha \vec{a}+\beta \vec{b} \right) \\
 & \Rightarrow {{\alpha }^{2}}\vec{a}\cdot \vec{a}+\beta \alpha \left( \vec{b}\cdot \vec{a} \right)-\alpha \left( \vec{a}\times \vec{b} \right)\cdot \vec{a}+\alpha \beta \vec{a}\cdot \vec{b}+{{\beta }^{2}}\vec{b}\cdot \vec{b}-\beta \left( \vec{a}\times \vec{b} \right)\cdot \vec{b} \\
 & \Rightarrow {{\alpha }^{2}}{{\left| {\vec{a}} \right|}^{2}}+\alpha \beta \left( \vec{a}\cdot \vec{b} \right)-\alpha \left[ \left( \vec{a}\times \vec{b} \right)\cdot \vec{a} \right]+\alpha \beta \left( \vec{a}\cdot \vec{b} \right)+{{\beta }^{2}}{{\left| {\vec{b}} \right|}^{2}}-\beta \left[ \left( \vec{a}\times \vec{b} \right)\cdot \vec{b} \right] \\
\end{align}\]
Using values of ${{\left| {\vec{a}} \right|}^{2}},{{\left| {\vec{b}} \right|}^{2}}\text{ and }\left( \vec{a}\cdot \vec{b} \right)$ we get:
\[\Rightarrow 6{{\alpha }^{2}}+3\alpha \beta -\alpha \left[ \left( \vec{a}\times \vec{b} \right)\cdot \vec{a} \right]+3\alpha \beta +6{{\beta }^{2}}-\beta \left[ \left( \vec{a}\times \vec{b} \right)\cdot \vec{b} \right]\]
As we can see $\left( \vec{a}\times \vec{b} \right)\cdot \vec{a}\text{ and }\left( \vec{a}\times \vec{b} \right)\cdot \vec{b}$ are triple products with two same vectors, so its value will be zero.
Hence, above equation becomes $\Rightarrow 6{{\alpha }^{2}}+6{{\beta }^{2}}+3\alpha \beta +3\alpha \beta =6\left[ {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right]$.
Let us add and subtract $2\alpha \beta $ in above, we get: $\Rightarrow 6\left( {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -2\alpha \beta +\alpha \beta \right)=6\left( {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -\alpha \beta \right)$
We know, ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$ using that, we get: $\Rightarrow 6\left( {{\left( \alpha +\beta \right)}^{2}}-\alpha \beta \right)$
Now, we know $\alpha +\beta =2$ therefore, we get $\Rightarrow 6\left[ 4-\alpha \beta \right]$.
$\beta $ can be written as $2-\alpha $ therefore,
\[\begin{align}
  & \Rightarrow 6\left[ 4-\alpha \left( 2-\alpha \right) \right] \\
 & \Rightarrow 6\left[ 4-2\alpha +{{\alpha }^{2}} \right]\ldots \ldots \ldots \left( 3 \right) \\
\end{align}\]
For calculating minimum value, let us suppose $f\left( \alpha \right)={{\alpha }^{2}}-2\alpha +4$.
Taking the derivative $f'\left( \alpha \right)=2\alpha -2$.
Putting $f'\left( \alpha \right)=0$ we get: $\Rightarrow 2\alpha -2=0\Rightarrow \alpha -1=0$
Therefore, $\alpha =1$ for checking if its maximum or minimum value, let us again take the derivative of $f''\left( \alpha \right)$ we get: $f''\left( \alpha \right)=2$. Putting $\alpha =1$ $f''\left( 1 \right)=2\text{ }>\text{ }0$.
Therefore by the second derivative test, $f\left( \alpha \right)$ is the minimum value when $\alpha =1$.
Hence, our minimum value becomes, putting $\alpha =1$in (3), we get:
$\Rightarrow 6\left( 4-2+1 \right)=18$.
Hence, the minimum value of $\left( \vec{c}-\left( \vec{a}\times \vec{b} \right) \right)\cdot \vec{c}$ equals 18.

Note: Students should take care of signs while solving problems related to vectors. While taking triple product of $\left( \vec{a}\times \vec{b} \right)\cdot \vec{c}$ if any two vector are same, the product becomes equal to zero because of the determinant property. Since, the triple product is calculated using determinant and if two rows become equal, then determinant is zero and hence, the scalar triple product is zero. It should be noted that $\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$ but $\vec{a}\times \vec{b}\ne \vec{b}\times \vec{a}$.