Let the vectors $\overrightarrow{a}=\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k}$, $\overrightarrow{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+\sqrt{2}\widehat{k}$ and $\overrightarrow{c}=5\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k}$ be three vectors such that the projection vector of $\overrightarrow{b}$on $\overrightarrow{a}$ is $\overrightarrow{a}$. If $\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$, then $\left| \overrightarrow{b} \right|$ is equal to:
(a) $\sqrt{22}$
(b) 4
(c) $\sqrt{34}$
(d) 6
Answer
Verified653.7k+ views
Hint: First, before proceeding for this, we must know the dot product of the two vectors is given by the formula as $\cos \theta =\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}$. Then, we will use the formula for the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ where $\widehat{a}$ is unit vector defined by $\widehat{a}=\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$ is given by $\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|}\widehat{a}$. Then, we are also given in the question that $\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$and by using it in the formula $\cos \theta =\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}$, we get the final result by using $\left| \overrightarrow{b} \right|=\sqrt{{{b}_{1}}^{2}+{{b}_{2}}^{2}}$.
Complete step-by-step solution:
In this question, we are supposed to find the value of $\left| \overrightarrow{b} \right|$ when $\overrightarrow{a}=\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k}$, $\overrightarrow{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+\sqrt{2}\widehat{k}$ and $\overrightarrow{c}=5\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k}$ be three vectors such that the projection vector of $\overrightarrow{b}$on $\overrightarrow{a}$ is $\overrightarrow{a}$ and $\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$.
So, before proceeding for this, we must know the dot product of the two vectors is given by the formula as:
$\cos \theta =\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}$
Then, we are given with the condition that the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\overrightarrow{a}$.
So, we will use the formula for the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ where $\widehat{a}$ is unit vector defined by $\widehat{a}=\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$ is given by:
$\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|}\widehat{a}$
So, by substituting the value of $\widehat{a}$ in above formula, we get:
$\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|}\times \dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$
Now, we are a=given in the question that the value of above expression is $\overrightarrow{a}$, so by equating it, we get:
$\begin{align}
& \dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|}\times \dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}=\overrightarrow{a} \\
& \Rightarrow \dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{{{\left| \overrightarrow{a} \right|}^{2}}}=1 \\
& \Rightarrow \overrightarrow{a}\centerdot \overrightarrow{b}={{\left| \overrightarrow{a} \right|}^{2}} \\
\end{align}$
Then, by substituting the values of vectors $\overrightarrow{b}$ and $\overrightarrow{a}$ given in the question, we get:
$\begin{align}
& \left( \widehat{i}+\widehat{j}+\sqrt{2}\widehat{k} \right)\centerdot \left( {{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+\sqrt{2}\widehat{k} \right)={{\left( \sqrt{{{1}^{2}}+{{1}^{2}}+{{\sqrt{2}}^{2}}} \right)}^{2}} \\
& \Rightarrow {{b}_{1}}+{{b}_{2}}+\sqrt{2}\times \sqrt{2}=1+1+2 \\
& \Rightarrow {{b}_{1}}+{{b}_{2}}+2=4 \\
& \Rightarrow {{b}_{1}}+{{b}_{2}}=2....\left( i \right) \\
\end{align}$
Then, we are also given in the question that $\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$ and by using it in the formula $\cos \theta =\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}$, we get:
$\begin{align}
& \dfrac{\left( \overrightarrow{a}+\overrightarrow{b} \right)\centerdot \overrightarrow{c}}{\left| \overrightarrow{a}+\overrightarrow{b} \right|\left| \overrightarrow{c} \right|}=\cos {{90}^{\circ }} \\
& \Rightarrow \left( \overrightarrow{a}+\overrightarrow{b} \right)\centerdot \overrightarrow{c}=0 \\
\end{align}$
Then, by substituting the value of all the vectors in the above expression, we get:
$\begin{align}
& \left( \widehat{i}+\widehat{j}+\sqrt{2}\widehat{k}+{{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+\sqrt{2}\widehat{k} \right)\centerdot \left( 5\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k} \right)=0 \\
& \Rightarrow \left( \left( 1+{{b}_{1}} \right)\widehat{i}+\left( 1+{{b}_{2}} \right)\widehat{j}+2\sqrt{2}\widehat{k} \right)\centerdot \left( 5\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k} \right)=0 \\
& \Rightarrow \left( \left( 1+{{b}_{1}} \right)5+\left( 1+{{b}_{2}} \right)1+2\sqrt{2}\times \sqrt{2} \right)=0 \\
& \Rightarrow 5+5{{b}_{1}}+1+{{b}_{2}}+4=0 \\
& \Rightarrow 5{{b}_{1}}+{{b}_{2}}=-10.....\left( ii \right) \\
\end{align}$
Now, by substituting the value of ${{b}_{1}}=2-{{b}_{2}}$ from equation (i) into equation (ii), we get:
$\begin{align}
& 5\left( 2-{{b}_{2}} \right)+{{b}_{2}}=-10 \\
& \Rightarrow 10-5{{b}_{2}}+{{b}_{2}}=-10 \\
& \Rightarrow -4{{b}_{2}}=-20 \\
& \Rightarrow {{b}_{2}}=\dfrac{-20}{-4} \\
& \Rightarrow {{b}_{2}}=5 \\
\end{align}$
Then, by substituting the value of ${{b}_{2}}$ as 5 in equation (i), we get:
$\begin{align}
& {{b}_{1}}+5=2 \\
& \Rightarrow {{b}_{1}}=-3 \\
\end{align}$
So, we need to get the value of magnitude of vector b is given by:
$\left| \overrightarrow{b} \right|=\sqrt{{{b}_{1}}^{2}+{{b}_{2}}^{2}}$
Now, by substituting the value of ${{b}_{1}}$as -3 and ${{b}_{2}}$as 5, we get:
$\begin{align}
& \left| \overrightarrow{b} \right|=\sqrt{{{\left( -3 \right)}^{2}}+{{5}^{2}}} \\
& \Rightarrow \left| \overrightarrow{b} \right|=\sqrt{9+25} \\
& \Rightarrow \left| \overrightarrow{b} \right|=\sqrt{34} \\
\end{align}$
So, we get the value of $\left| \overrightarrow{b} \right|$ as$\sqrt{34}$.
Hence, option (c) is correct.
Note: Now, to solve these types of the questions we need to know some of the other methods to solve the two-variable equations. So, the other two methods are the elimination method and cross multiplication method to get the answers of ${{b}_{1}}$ and ${{b}_{2}}$.
Complete step-by-step solution:
In this question, we are supposed to find the value of $\left| \overrightarrow{b} \right|$ when $\overrightarrow{a}=\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k}$, $\overrightarrow{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+\sqrt{2}\widehat{k}$ and $\overrightarrow{c}=5\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k}$ be three vectors such that the projection vector of $\overrightarrow{b}$on $\overrightarrow{a}$ is $\overrightarrow{a}$ and $\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$.
So, before proceeding for this, we must know the dot product of the two vectors is given by the formula as:
$\cos \theta =\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}$
Then, we are given with the condition that the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\overrightarrow{a}$.
So, we will use the formula for the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ where $\widehat{a}$ is unit vector defined by $\widehat{a}=\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$ is given by:
$\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|}\widehat{a}$
So, by substituting the value of $\widehat{a}$ in above formula, we get:
$\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|}\times \dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$
Now, we are a=given in the question that the value of above expression is $\overrightarrow{a}$, so by equating it, we get:
$\begin{align}
& \dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|}\times \dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}=\overrightarrow{a} \\
& \Rightarrow \dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{{{\left| \overrightarrow{a} \right|}^{2}}}=1 \\
& \Rightarrow \overrightarrow{a}\centerdot \overrightarrow{b}={{\left| \overrightarrow{a} \right|}^{2}} \\
\end{align}$
Then, by substituting the values of vectors $\overrightarrow{b}$ and $\overrightarrow{a}$ given in the question, we get:
$\begin{align}
& \left( \widehat{i}+\widehat{j}+\sqrt{2}\widehat{k} \right)\centerdot \left( {{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+\sqrt{2}\widehat{k} \right)={{\left( \sqrt{{{1}^{2}}+{{1}^{2}}+{{\sqrt{2}}^{2}}} \right)}^{2}} \\
& \Rightarrow {{b}_{1}}+{{b}_{2}}+\sqrt{2}\times \sqrt{2}=1+1+2 \\
& \Rightarrow {{b}_{1}}+{{b}_{2}}+2=4 \\
& \Rightarrow {{b}_{1}}+{{b}_{2}}=2....\left( i \right) \\
\end{align}$
Then, we are also given in the question that $\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$ and by using it in the formula $\cos \theta =\dfrac{\overrightarrow{a}\centerdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}$, we get:
$\begin{align}
& \dfrac{\left( \overrightarrow{a}+\overrightarrow{b} \right)\centerdot \overrightarrow{c}}{\left| \overrightarrow{a}+\overrightarrow{b} \right|\left| \overrightarrow{c} \right|}=\cos {{90}^{\circ }} \\
& \Rightarrow \left( \overrightarrow{a}+\overrightarrow{b} \right)\centerdot \overrightarrow{c}=0 \\
\end{align}$
Then, by substituting the value of all the vectors in the above expression, we get:
$\begin{align}
& \left( \widehat{i}+\widehat{j}+\sqrt{2}\widehat{k}+{{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+\sqrt{2}\widehat{k} \right)\centerdot \left( 5\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k} \right)=0 \\
& \Rightarrow \left( \left( 1+{{b}_{1}} \right)\widehat{i}+\left( 1+{{b}_{2}} \right)\widehat{j}+2\sqrt{2}\widehat{k} \right)\centerdot \left( 5\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k} \right)=0 \\
& \Rightarrow \left( \left( 1+{{b}_{1}} \right)5+\left( 1+{{b}_{2}} \right)1+2\sqrt{2}\times \sqrt{2} \right)=0 \\
& \Rightarrow 5+5{{b}_{1}}+1+{{b}_{2}}+4=0 \\
& \Rightarrow 5{{b}_{1}}+{{b}_{2}}=-10.....\left( ii \right) \\
\end{align}$
Now, by substituting the value of ${{b}_{1}}=2-{{b}_{2}}$ from equation (i) into equation (ii), we get:
$\begin{align}
& 5\left( 2-{{b}_{2}} \right)+{{b}_{2}}=-10 \\
& \Rightarrow 10-5{{b}_{2}}+{{b}_{2}}=-10 \\
& \Rightarrow -4{{b}_{2}}=-20 \\
& \Rightarrow {{b}_{2}}=\dfrac{-20}{-4} \\
& \Rightarrow {{b}_{2}}=5 \\
\end{align}$
Then, by substituting the value of ${{b}_{2}}$ as 5 in equation (i), we get:
$\begin{align}
& {{b}_{1}}+5=2 \\
& \Rightarrow {{b}_{1}}=-3 \\
\end{align}$
So, we need to get the value of magnitude of vector b is given by:
$\left| \overrightarrow{b} \right|=\sqrt{{{b}_{1}}^{2}+{{b}_{2}}^{2}}$
Now, by substituting the value of ${{b}_{1}}$as -3 and ${{b}_{2}}$as 5, we get:
$\begin{align}
& \left| \overrightarrow{b} \right|=\sqrt{{{\left( -3 \right)}^{2}}+{{5}^{2}}} \\
& \Rightarrow \left| \overrightarrow{b} \right|=\sqrt{9+25} \\
& \Rightarrow \left| \overrightarrow{b} \right|=\sqrt{34} \\
\end{align}$
So, we get the value of $\left| \overrightarrow{b} \right|$ as$\sqrt{34}$.
Hence, option (c) is correct.
Note: Now, to solve these types of the questions we need to know some of the other methods to solve the two-variable equations. So, the other two methods are the elimination method and cross multiplication method to get the answers of ${{b}_{1}}$ and ${{b}_{2}}$.
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