Question

The scalar product of the vector $\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}$ with a unit vector along the sum of vectors $\overrightarrow{b}=2\widehat{i}+4\widehat{j}-5\widehat{k}$ and $\overrightarrow{c}=\lambda \widehat{i}+2\widehat{j}+3\widehat{k}$ is equal to one. Find the value of $\lambda $ and hence find the unit vector along $\overrightarrow{b}+\overrightarrow{c}$.

Answer 1

Hint: First, before proceeding for this, we must know the concept that in vector addition and subtraction only like components get added or subtracted to get the value of vector$\overrightarrow{b}+\overrightarrow{c}$. Then, let us suppose that n be the unit vector along $\overrightarrow{b}+\overrightarrow{c}$ and the value of the unit vector n is given by $\widehat{n}=\dfrac{\overrightarrow{b}+\overrightarrow{c}}{\left| \overrightarrow{b}+\overrightarrow{c} \right|}$. Then, in the question, we are given the condition that the above unit vector n and the vector a as scalar product as 1 to get the value of $\lambda $. Then, by substituting the value of $\lambda $ in the unit n vector to get its value.

Complete step-by-step answer:
In this question, we are supposed to find the value of $\lambda $from the condition given that the scalar product of the vector $\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}$ with a unit vector along the sum of vectors $\overrightarrow{b}=2\widehat{i}+4\widehat{j}-5\widehat{k}$ and $\overrightarrow{c}=\lambda \widehat{i}+2\widehat{j}+3\widehat{k}$ is equal to one and also find the unit vector along $\overrightarrow{b}+\overrightarrow{c}$.
So, before proceeding for this, we must know the concept that in vector addition and subtraction only like components get added or subtracted.
Then, by using the above concept, we need to find the value of the summation of vectors of $\overrightarrow{b}+\overrightarrow{c}$as:
$\overrightarrow{b}+\overrightarrow{c}=2\widehat{i}+4\widehat{j}-5\widehat{k}+\lambda \widehat{i}+2\widehat{j}+3\widehat{k}$
So, by solving the above expression, we get:
$\overrightarrow{b}+\overrightarrow{c}=\left( 2+\lambda \right)\widehat{i}+6\widehat{j}-2\widehat{k}$
Now, let us suppose that n be the unit vector along $\overrightarrow{b}+\overrightarrow{c}$ and the value of the unit vector n is given by:
$\widehat{n}=\dfrac{\overrightarrow{b}+\overrightarrow{c}}{\left| \overrightarrow{b}+\overrightarrow{c} \right|}$
Then, by solving for the unit vector n, we get:
$\begin{align}
  & \widehat{n}=\dfrac{\left( 2+\lambda \right)\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{{{\left( 2+\lambda \right)}^{2}}+{{6}^{2}}+{{\left( -2 \right)}^{2}}}} \\
 & \Rightarrow \widehat{n}=\dfrac{\left( 2+\lambda \right)\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{{{\lambda }^{2}}+4+4\lambda +36+4}} \\
 & \Rightarrow \widehat{n}=\dfrac{\left( 2+\lambda \right)\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{{{\lambda }^{2}}+4\lambda +44}} \\
\end{align}$
Now, in the question, we are given with the condition that the above unit vector n and the vector a as scalar product as 1.
So, by using this condition, we get:
$\overrightarrow{a}\centerdot \widehat{n}=1$
Now, by substituting the value of vector a and unit vector n calculated above, we get:
$\left( \widehat{i}+\widehat{j}+\widehat{k} \right)\centerdot \left( \dfrac{\left( 2+\lambda \right)\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{{{\lambda }^{2}}+4\lambda +44}} \right)=1$
Then, by solving the above expression, we get:
$\begin{align}
  & \left( 2+\lambda \right)+6-2=\sqrt{{{\lambda }^{2}}+4\lambda +44} \\
 & \Rightarrow \lambda +6=\sqrt{{{\lambda }^{2}}+4\lambda +44} \\
\end{align}$
Then, by squaring both sides, we get:
$\begin{align}
  & {{\left( \lambda +6 \right)}^{2}}={{\lambda }^{2}}+4\lambda +44 \\
 & \Rightarrow {{\lambda }^{2}}+12\lambda +36={{\lambda }^{2}}+4\lambda +44 \\
 & \Rightarrow 8\lambda =8 \\
 & \Rightarrow \lambda =1 \\
\end{align}$
So, we get the value of $\lambda $ as 1.
Now, by substituting the value of $\lambda $ as 1 in unit n vector to get its value as:
$\begin{align}
  & \widehat{n}=\dfrac{\left( 2+1 \right)\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{{{1}^{2}}+4\left( 1 \right)+44}} \\
 & \Rightarrow \widehat{n}=\dfrac{3\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{1+4+44}} \\
 & \Rightarrow \widehat{n}=\dfrac{3\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{49}} \\
 & \Rightarrow \widehat{n}=\dfrac{3\widehat{i}+6\widehat{j}-2\widehat{k}}{7} \\
\end{align}$
So, we get the value of unit vector along $\overrightarrow{b}+\overrightarrow{c}$ as $\widehat{n}=\dfrac{3\widehat{i}+6\widehat{j}-2\widehat{k}}{7}$.


Note: To solve these types of questions we need to know some of the basic calculations of the dot product or scalar product. So, if we have two vectors as $X\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)$ and $Y\left( d\widehat{i}+e\widehat{j}+f\widehat{k} \right)$, then their scalar product is given by:
$X\centerdot Y=a\times d+b\times e+c\times f$.