Question

Find a vector $\overrightarrow{p}$ such that it is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$, and $\overrightarrow{p}.\overrightarrow{c}=18$, where $\overrightarrow{a}=\hat{i}+4\hat{j}+2\hat{k}$, $\overrightarrow{b}=3\hat{i}-2\hat{j}+7\hat{k}$ and $\overrightarrow{c}=2\hat{i}-\hat{j}+4\hat{k}$.

Answer 1

We solve this question by finding the vector perpendicular to given two vectors using the formula $\left( \overrightarrow{a}\times \overrightarrow{b} \right)=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ \end{matrix} \right|$. Then we multiply the obtained vector with some constant and consider it as $\overrightarrow{p}$. Then we find the dot product of $\overrightarrow{p}$ and $\overrightarrow{c}$ using the formula for dot product$~\overrightarrow{a}.\overrightarrow{b}={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}$. Then we can find the value of the constant and substitute it in the value vector $\overrightarrow{p}$ to find the value of the vector $\overrightarrow{p}$.

Complete step by step answer:
For solving this problem, we need to go through the concept of dot product of two vectors.
Dot product of vectors $\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and $\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ can be given as,
$~\overrightarrow{a}.\overrightarrow{b}={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}$
Any vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$ can be written in the form of $k\left( \overrightarrow{a}\times \overrightarrow{b} \right)$, where k is a constant. While $\left( \overrightarrow{a}\times \overrightarrow{b} \right)$ is the cross product of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ can be given as,
$\left( \overrightarrow{a}\times \overrightarrow{b} \right)=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ \end{matrix} \right|$
We are given that a vector $\overrightarrow{p}$ is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$, and $\overrightarrow{p}.\overrightarrow{c}=18$.
As $\overrightarrow{p}$ is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$, we can write $\overrightarrow{p}$ as $l\left( \overrightarrow{a}\times \overrightarrow{b} \right)$, where l is some constant.
First let us calculate the value of $\left( \overrightarrow{a}\times \overrightarrow{b} \right)$.
$\begin{aligned} & \Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 1 & 4 & 2 \\\ 3 & -2 & 7 \\\ \end{matrix} \right| \\\ & \Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)=\hat{i}\left| \begin{matrix} 4 & 2 \\\ -2 & 7 \\\ \end{matrix} \right|-\hat{j}\left| \begin{matrix} 1 & 2 \\\ 3 & 7 \\\ \end{matrix} \right|+\hat{k}\left| \begin{matrix} 1 & 4 \\\ 3 & -2 \\\ \end{matrix} \right| \\\ & \Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)=\hat{i}\left( 28+4 \right)-\hat{j}\left( 7-6 \right)+\hat{k}\left( -2-12 \right) \\\ & \Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)=32\hat{i}-\hat{j}-14\hat{k} \\\ \end{aligned}$
As $\overrightarrow{p}$ is in the form of $l\left( \overrightarrow{a}\times \overrightarrow{b} \right)$, we can write $\overrightarrow{p}$ as
$\begin{aligned} & \Rightarrow \overrightarrow{p}=l\left( \overrightarrow{a}\times \overrightarrow{b} \right) \\\ & \Rightarrow \overrightarrow{p}=l\left( 32\hat{i}-\hat{j}-14\hat{k} \right) \\\ & \Rightarrow \overrightarrow{p}=32l\hat{i}-l\hat{j}-14l\hat{k} \\\ \end{aligned}$
We are given that $\overrightarrow{p}.\overrightarrow{c}=18$.
So, by using the above formula of dot product of $\overrightarrow{p}$ and $\overrightarrow{c}$, we get

& \Rightarrow \overrightarrow{p}.\overrightarrow{c}=18 \\\ & \Rightarrow ~\left( 32l \right)\left( 2 \right)+\left( -l \right)\left( -1 \right)+\left( -14l \right)\left( 4 \right)=18 \\\ & \Rightarrow ~64l+l-56l=18 \\\ & \Rightarrow ~9l=18 \\\ & \Rightarrow ~l=\dfrac{18}{9} \\\ & \Rightarrow ~l=2 \\\ \end{aligned}$$ Then by substituting the value of $l$ in $\overrightarrow{p}$, we get $\begin{aligned} & \Rightarrow \overrightarrow{p}=32\left( 2 \right)\hat{i}-\left( 2 \right)\hat{j}-14\left( 2 \right)\hat{k} \\\ & \Rightarrow \overrightarrow{p}=64\hat{i}-2\hat{j}-28\hat{k} \\\ \end{aligned}$ So, the value of vector $\overrightarrow{p}$ is $\overrightarrow{p}=64\hat{i}-2\hat{j}-28\hat{k}$. Hence the answer is $\overrightarrow{p}=64\hat{i}-2\hat{j}-28\hat{k}$. **Note:** The major mistake that one makes while solving this problem is exchanging the formulas of cross product and dot product for one another like considering $\overrightarrow{a}.\overrightarrow{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ \end{matrix} \right|$ and $$~\overrightarrow{a}\times \overrightarrow{b}={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}$$, which is wrong. So, one needs to remember the formulas for dot product and cross product and use them correctly.