Let we have the vectors $\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$ , $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$ and $\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$. Find a vector $\vec{d}$ which is perpendicular to both $\vec{a},\vec{b}$ also $\vec{c}\cdot \vec{d}=15.$\[\]
Answer
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Hint: Take $\vec{d}$ as ${{d}_{1}}\hat{i}+{{d}_{2}}\hat{j}+{{d}_{3}}\hat{k}$. Use the fact that two vectors which are perpendicular, sum of product of respective components will be zero $\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}=0 \right)$ for given $\vec{a}$ and $\vec{b}$ . Use $\vec{c}\cdot \vec{d}=15.$ You will obtain $3\times 3$ system of linear equations in ${{d}_{1}},{{d}_{2}},{{d}_{3}}.$ Solve it by Kramer’s rule to get $\vec{d}$.\[\]
Complete step-by-step solution:
We know that the dot product of two vectors $\vec{a}$ and $\vec{b}$ is denoted as $\vec{a}\cdot \vec{b}$ and is given by $\vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta =ab\cos \theta $ where $\theta $ is the angle between the vectors $\vec{a}$ and $\vec{b}$. If they are perpendicular r to each other $\vec{a}\cdot \vec{b}=ab\cos {{90}^{\circ }}=0$\[\]
We also know that $\hat{i}$,$\hat{j}$ and $\hat{k}$ are orthogonal unit vectors(vectors with magnitude 1) along $x,y$ and $z$ axes respectively. So the magnitude of these vectors is $\left| {\hat{i}} \right|=\left| {\hat{j}} \right|=\left| {\hat{k}} \right|=1 $. The vectors just like their axes are perpendicular to each other which mean any angle among$\hat{i}$,$\hat{j}$ and $\hat{k}$is ${{90}^{\circ }}.$ So $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1\cdot 1\cdot \cos {{0}^{\circ }}=1$ and $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{i}=\hat{k}\cdot \hat{i}=0$\[\]
We can express any vector as a linear combination of orthogonal unit vectors. If $\vec{a}$ is expressed as $\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and $\vec{b}$ is expressed as $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ where ${{a}_{1}},{{a}_{2}},{{a}_{3}}$ are the magnitude of components of the vector $\vec{a}$ and ${{b}_{1}},{{b}_{2}},{{b}_{3}}$ are the magnitude of components the vector $\vec{b}$ along the direction of unit orthogonal vectors $\hat{i}$,$\hat{j}$ and $\hat{k}$ . If $\vec{a}$ and $\vec{b}$ are going orthogonal when
\[\begin{align}
& \vec{a}\cdot \vec{b}=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right)\cdot \left( {{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right)=0 \\
& \Rightarrow {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}=0...(1) \\
\end{align}\]
We know the $3\times 3$ linear system of equations with three unknowns $x,y,z$ in matrix from as,
\[\left[ \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
{{d}_{1}} \\
{{d}_{2}} \\
{{d}_{2}} \\
\end{matrix} \right]\]
The unique solution of the above system is given by $x=\dfrac{{{\Delta }_{a}}}{\Delta },y=\dfrac{{{\Delta }_{b}}}{\Delta }x=\dfrac{{{\Delta }_{c}}}{\Delta }$ where $\Delta $ is determinant of the coefficient matrix
\[\Delta =\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\]
The other determinants are
\[{{\Delta }_{a}}=\left| \begin{matrix}
{{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|,{{\Delta }_{b}}=\left| \begin{matrix}
{{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\
\end{matrix} \right|,{{\Delta }_{c}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\
\end{matrix} \right|\]
Let us take the vector $\vec{d}$ as ${{d}_{1}}\hat{i}+{{d}_{2}}\hat{j}+{{d}_{3}}\hat{k}$. It is given in the question that $\vec{d}$ is perpendicular to $\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$ . So
\[{{d}_{1}}+4{{d}_{2}}+2{{d}_{3}}=0...(1)\]
It is given that the vector $\vec{d}$ is perpendicular $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$ . So
\[3{{d}_{1}}-2{{d}_{2}}+7{{d}_{3}}=0...(2)\]
It is also given that $\vec{c}\cdot \vec{d}=15.$ So
\[ 3{{d}_{1}}-{{d}_{2}}+4{{d}_{3}}=15...(3)\]
The obtained equations (1) (2) and (3) form a linear system of equations with three unknowns
The system of equations in the matrix form is ,
\[\left[ \begin{matrix}
1 & 4 & 2 \\
3 & -2 & 7 \\
2 & -1 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
{{d}_{1}} \\
{{d}_{2}} \\
{{d}_{3}} \\
\end{matrix} \right]=\left[ \begin{matrix}
0 \\
0 \\
15 \\
\end{matrix} \right]\]
We solve the above system of equations with Cramer’s rule. We first find the determinant of the coefficient matrix
\[\Delta =\left| \begin{matrix}
1 & 4 & 2 \\
3 & -2 & 7 \\
2 & -1 & 4 \\
\end{matrix} \right|=1\left( -8+7 \right)-4\left( 12-14 \right)+2\left( -3+4 \right)=9\]
We find the other determinants
\[\begin{align}
& {{\Delta }_{a}}=\left| \begin{matrix}
0 & 4 & 2 \\
0 & -2 & 7 \\
15 & -1 & 4 \\
\end{matrix} \right|=15\left( 16-\left( -2 \right) \right)=480, \\
& {{\Delta }_{b}}=\left| \begin{matrix}
1 & 0 & 2 \\
3 & 0 & 7 \\
2 & 15 & 4 \\
\end{matrix} \right|=-15\left( 7-6 \right)=-15, \\
& {{\Delta }_{c}}=\left| \begin{matrix}
1 & 4 & 0 \\
3 & -2 & 0 \\
2 & -1 & 15 \\
\end{matrix} \right|=15\left( -2-12 \right)=-210 \\
\end{align}\]
So the solutions of the equations are
\[\begin{align}
& {{d}_{1}}=\dfrac{{{\Delta }_{a}}}{\Delta }=\dfrac{480}{9}=\dfrac{160}{3} \\
& {{d}_{2}}=\dfrac{{{\Delta }_{b}}}{\Delta }=\dfrac{-15}{9}=\dfrac{-5}{3} \\
& {{d}_{3}}=\dfrac{{{\Delta }_{c}}}{\Delta }=\dfrac{-210}{9}=\dfrac{-70}{3} \\
\end{align}\]
So the vector $\vec{d}$ is $\vec{d}=\dfrac{160}{3}\hat{i}-\dfrac{5}{3}\hat{j}-\dfrac{70}{3}\hat{k}$\[\]
Note: The perpendicular condition for two vectors with $n$ component is given as ${{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+...+{{a}_{n}}{{b}_{n}}=0$. We can also taking $\vec{d}=k\left( \vec{a}\times \vec{b} \right)$ where $k$ is a real number and $\vec{a}\times \vec{b}$ is the cross product. Then we use $\vec{d}$ in $\vec{c}\cdot \vec{d}=15$ to find $k$. We should note that dot product is commutative but cross-product is not.
Complete step-by-step solution:
We know that the dot product of two vectors $\vec{a}$ and $\vec{b}$ is denoted as $\vec{a}\cdot \vec{b}$ and is given by $\vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta =ab\cos \theta $ where $\theta $ is the angle between the vectors $\vec{a}$ and $\vec{b}$. If they are perpendicular r to each other $\vec{a}\cdot \vec{b}=ab\cos {{90}^{\circ }}=0$\[\]
We also know that $\hat{i}$,$\hat{j}$ and $\hat{k}$ are orthogonal unit vectors(vectors with magnitude 1) along $x,y$ and $z$ axes respectively. So the magnitude of these vectors is $\left| {\hat{i}} \right|=\left| {\hat{j}} \right|=\left| {\hat{k}} \right|=1 $. The vectors just like their axes are perpendicular to each other which mean any angle among$\hat{i}$,$\hat{j}$ and $\hat{k}$is ${{90}^{\circ }}.$ So $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1\cdot 1\cdot \cos {{0}^{\circ }}=1$ and $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{i}=\hat{k}\cdot \hat{i}=0$\[\]
We can express any vector as a linear combination of orthogonal unit vectors. If $\vec{a}$ is expressed as $\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and $\vec{b}$ is expressed as $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ where ${{a}_{1}},{{a}_{2}},{{a}_{3}}$ are the magnitude of components of the vector $\vec{a}$ and ${{b}_{1}},{{b}_{2}},{{b}_{3}}$ are the magnitude of components the vector $\vec{b}$ along the direction of unit orthogonal vectors $\hat{i}$,$\hat{j}$ and $\hat{k}$ . If $\vec{a}$ and $\vec{b}$ are going orthogonal when
\[\begin{align}
& \vec{a}\cdot \vec{b}=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right)\cdot \left( {{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right)=0 \\
& \Rightarrow {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}=0...(1) \\
\end{align}\]
We know the $3\times 3$ linear system of equations with three unknowns $x,y,z$ in matrix from as,
\[\left[ \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
{{d}_{1}} \\
{{d}_{2}} \\
{{d}_{2}} \\
\end{matrix} \right]\]
The unique solution of the above system is given by $x=\dfrac{{{\Delta }_{a}}}{\Delta },y=\dfrac{{{\Delta }_{b}}}{\Delta }x=\dfrac{{{\Delta }_{c}}}{\Delta }$ where $\Delta $ is determinant of the coefficient matrix
\[\Delta =\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\]
The other determinants are
\[{{\Delta }_{a}}=\left| \begin{matrix}
{{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|,{{\Delta }_{b}}=\left| \begin{matrix}
{{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\
\end{matrix} \right|,{{\Delta }_{c}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\
\end{matrix} \right|\]
Let us take the vector $\vec{d}$ as ${{d}_{1}}\hat{i}+{{d}_{2}}\hat{j}+{{d}_{3}}\hat{k}$. It is given in the question that $\vec{d}$ is perpendicular to $\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$ . So
\[{{d}_{1}}+4{{d}_{2}}+2{{d}_{3}}=0...(1)\]
It is given that the vector $\vec{d}$ is perpendicular $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$ . So
\[3{{d}_{1}}-2{{d}_{2}}+7{{d}_{3}}=0...(2)\]
It is also given that $\vec{c}\cdot \vec{d}=15.$ So
\[ 3{{d}_{1}}-{{d}_{2}}+4{{d}_{3}}=15...(3)\]
The obtained equations (1) (2) and (3) form a linear system of equations with three unknowns
The system of equations in the matrix form is ,
\[\left[ \begin{matrix}
1 & 4 & 2 \\
3 & -2 & 7 \\
2 & -1 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
{{d}_{1}} \\
{{d}_{2}} \\
{{d}_{3}} \\
\end{matrix} \right]=\left[ \begin{matrix}
0 \\
0 \\
15 \\
\end{matrix} \right]\]
We solve the above system of equations with Cramer’s rule. We first find the determinant of the coefficient matrix
\[\Delta =\left| \begin{matrix}
1 & 4 & 2 \\
3 & -2 & 7 \\
2 & -1 & 4 \\
\end{matrix} \right|=1\left( -8+7 \right)-4\left( 12-14 \right)+2\left( -3+4 \right)=9\]
We find the other determinants
\[\begin{align}
& {{\Delta }_{a}}=\left| \begin{matrix}
0 & 4 & 2 \\
0 & -2 & 7 \\
15 & -1 & 4 \\
\end{matrix} \right|=15\left( 16-\left( -2 \right) \right)=480, \\
& {{\Delta }_{b}}=\left| \begin{matrix}
1 & 0 & 2 \\
3 & 0 & 7 \\
2 & 15 & 4 \\
\end{matrix} \right|=-15\left( 7-6 \right)=-15, \\
& {{\Delta }_{c}}=\left| \begin{matrix}
1 & 4 & 0 \\
3 & -2 & 0 \\
2 & -1 & 15 \\
\end{matrix} \right|=15\left( -2-12 \right)=-210 \\
\end{align}\]
So the solutions of the equations are
\[\begin{align}
& {{d}_{1}}=\dfrac{{{\Delta }_{a}}}{\Delta }=\dfrac{480}{9}=\dfrac{160}{3} \\
& {{d}_{2}}=\dfrac{{{\Delta }_{b}}}{\Delta }=\dfrac{-15}{9}=\dfrac{-5}{3} \\
& {{d}_{3}}=\dfrac{{{\Delta }_{c}}}{\Delta }=\dfrac{-210}{9}=\dfrac{-70}{3} \\
\end{align}\]
So the vector $\vec{d}$ is $\vec{d}=\dfrac{160}{3}\hat{i}-\dfrac{5}{3}\hat{j}-\dfrac{70}{3}\hat{k}$\[\]
Note: The perpendicular condition for two vectors with $n$ component is given as ${{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+...+{{a}_{n}}{{b}_{n}}=0$. We can also taking $\vec{d}=k\left( \vec{a}\times \vec{b} \right)$ where $k$ is a real number and $\vec{a}\times \vec{b}$ is the cross product. Then we use $\vec{d}$ in $\vec{c}\cdot \vec{d}=15$ to find $k$. We should note that dot product is commutative but cross-product is not.
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