Question

let \[\overset{\to }{\mathop{a}}\,=\widehat{i}+4\widehat{j}+2\widehat{k},\text{ }\overset{\to }{\mathop{b}}\,=3\widehat{i}-2\widehat{j}-7\widehat{k}\text{ and }\overset{\to }{\mathop{c}}\,=2\widehat{i}-\widehat{j}+4\widehat{k}\text{.}\] find a vector\[\overset{\to }{\mathop{p}}\,\]which is
perpendicular to both $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ and $\overset{\to }{\mathop{p}}\,\cdot \overset{\to }{\mathop{c}}\,=18.$

Answer 1

Hint: In this problem we will find the perpendicular vector$\overset{\to }{\mathop{p}}\,$. If two vectors $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ are perpendicular then the dot product of two vectors $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ is zero.
i.e. $\overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{b}}\,=0$

Complete step by step answer:
Before, start solving the problem let us define a vector. A vector is a quantity which can be completely described using both magnitude and direction.
Mathematically, A vector is a line segment AB with direction from A to B and denoted by $\overset{\to }{\mathop{\text{AB}}}\,$
Given that \[\overset{\to }{\mathop{a}}\,=\widehat{i}+4\widehat{j}+2\widehat{k},\text{ }\overset{\to }{\mathop{b}}\,=3\widehat{i}-2\widehat{j}-7\widehat{k}\text{ and }\overset{\to }{\mathop{c}}\,=2\widehat{i}-\widehat{j}+4\widehat{k}\]
To find the vector \[\overset{\to }{\mathop{p}}\,\].
Let $\overset{\to }{\mathop{p}}\,={{p}_{1}}\widehat{i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k}$ be a vector perpendicular to both $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$and $\overset{\to }{\mathop{p}}\,\cdot \overset{\to }{\mathop{c}}\,=18$
Since $\overset{\to }{\mathop{p}}\,=\widehat{{{p}_{1}}i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k}$ is perpendicular to\[\overset{\to }{\mathop{a}}\,=\widehat{i}+4\widehat{j}+2\widehat{k}\]
$\Rightarrow $ Dot product of two vectors $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{p}}\,$ is zero
$\Rightarrow \overset{\to }{\mathop{p}}\,\cdot \overset{\to }{\mathop{a}}\,=0$
Now, we will substitute the vectors$\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{p}}\,$.
\[\Rightarrow \left( {{p}_{1}}\widehat{i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k} \right)\cdot \left( \widehat{i}+4\widehat{j}+2\widehat{k} \right)=0\]
By dot product of vectors, we get
\[\Rightarrow {{p}_{1}}+4{{p}_{2}}+2{{p}_{3}}=0....(1)\]
Since $\overset{\to }{\mathop{p}}\,=\widehat{{{p}_{1}}i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k}$ is perpendicular to\[\overset{\to }{\mathop{b}}\,=3\widehat{i}-2\widehat{j}-7\widehat{k}\]
$\Rightarrow $ dot product of two vectors $\overset{\to }{\mathop{b}}\,\text{ and }\overset{\to }{\mathop{p}}\,$ is zero
$\Rightarrow \overset{\to }{\mathop{p}}\,\cdot \overset{\to }{\mathop{b}}\,=0$
Now, we will substitute the vectors$\overset{\to }{\mathop{b}}\,\text{ and }\overset{\to }{\mathop{p}}\,$.
\[\Rightarrow \left( {{p}_{1}}\widehat{i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k} \right)\cdot \left( 3\widehat{i}-2\widehat{j}-7\widehat{k} \right)=0\]
By dot product of vectors, we get
\[\Rightarrow 3{{p}_{1}}-2{{p}_{2}}-7{{p}_{3}}=0....(2)\]
Also give that $\overset{\to }{\mathop{p}}\,\cdot \overset{\to }{\mathop{c}}\,=18$
Now, we will substitute the vectors$\overset{\to }{\mathop{c}}\,\text{ and }\overset{\to }{\mathop{p}}\,$.
\[\Rightarrow \left( {{p}_{1}}\widehat{i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k} \right)\cdot \left( 2\widehat{i}-\widehat{j}+4\widehat{k} \right)=18\]
By dot product of vectors, we get
\[\Rightarrow 2{{p}_{1}}-{{p}_{2}}+4{{p}_{3}}=18....(3)\]
From equation (1) we get
Using the value of \[{{p}_{1}}\] in equation (2),
\[\Rightarrow 3\left( -4{{p}_{2}}-2{{p}_{3}} \right)-2{{p}_{2}}-7{{p}_{3}}=0\]
\[\Rightarrow -12{{p}_{2}}-6{{p}_{3}}-2{{p}_{2}}-7{{p}_{3}}=0\]
\[\Rightarrow -14{{p}_{2}}-11{{p}_{3}}=0\]
\[\Rightarrow {{p}_{2}}=-\dfrac{11}{14}{{p}_{3}}....(5)\]
Using equation (5) in equation (4) we get,
\[\Rightarrow {{p}_{1}}=-4\left( -\dfrac{11}{14}{{p}_{3}} \right)-2{{p}_{3}}\]
\[\Rightarrow {{p}_{1}}=\dfrac{44}{14}{{p}_{3}}-2{{p}_{3}}\]
By cross multiplication we get
\[\Rightarrow {{p}_{1}}=\dfrac{44{{p}_{3}}-28{{p}_{3}}}{14}\]
\[\Rightarrow {{p}_{1}}=\dfrac{16{{p}_{3}}}{14}....(6)\]
Using equation (5) and equation (6) in equation (3), we get
\[\Rightarrow 2\left( \dfrac{16{{p}_{3}}}{14} \right)-\left( -\dfrac{11}{14}{{p}_{3}} \right)+4{{p}_{3}}=18....(3)\]
\[\Rightarrow \dfrac{32{{p}_{3}}}{14}+\dfrac{11}{14}{{p}_{3}}+4{{p}_{3}}=18\]
By cross multiplication, we get
\[\Rightarrow \dfrac{32{{p}_{3}}+11{{p}_{3}}+56{{p}_{3}}}{14}=18\]
\[\Rightarrow \dfrac{99{{p}_{3}}}{14}=18\]
By dividing LHS and RHS by 9, we get
\[\Rightarrow \dfrac{11{{p}_{3}}}{14}=2\]
By cross multiplication, we get
\[\Rightarrow {{p}_{3}}=\dfrac{28}{11}....(7)\]
Using equation (7) in equation (5) and equation (6) we get
\[\Rightarrow {{p}_{1}}=\dfrac{16}{14}\left( \dfrac{28}{11} \right)\]
\[\Rightarrow {{p}_{1}}=\dfrac{8}{7}\left( \dfrac{28}{11} \right)\]
\[\Rightarrow {{p}_{1}}=\dfrac{8\times 4}{11}\]
\[\Rightarrow {{p}_{1}}=\dfrac{32}{11}\]
And \[{{p}_{2}}=-\dfrac{11}{14}\left( \dfrac{28}{11} \right)\]
\[{{p}_{2}}=- 2\]
\[{{p}_{2}}=-2\]
Hence\[{{p}_{1}}=\dfrac{32}{11}\], \[{{p}_{2}}=-2\] and\[{{p}_{3}}=\dfrac{28}{11}\].
Therefore perpendicular vector is $\overset{\to }{\mathop{p}}\,=\dfrac{32}{11}\widehat{i}- 2\widehat{j}+\dfrac{28}{11}\widehat{k}$

Note:
In this problem, one knows that if two vectors are perpendicular then their dot product is zero. Alternative, to find the coordinate of vector $\overset{\to }{\mathop{p}}\,$ i.e. \[{{\text{p}}_{1}}\text{, }{{\text{p}}_{\text{2}}}\text{ and }{{\text{p}}_{3}}\] we can solve equation (1), equation (2) and equation (3) using matrices. By converting the system of equations in matrix form.