Question

Let $\overrightarrow{a}=4\widehat{i}+5\widehat{j}-\widehat{k},\overrightarrow{b}=\widehat{i}-4\widehat{j}+5\widehat{k}$ and $\overrightarrow{c}=3\widehat{i}+\widehat{j}-\widehat{k}$. Find a vector $\overrightarrow{d}$ which is perpendicular to both $\overrightarrow{c}$ and $\overrightarrow{b}$ and $\overrightarrow{d\cdot }\overrightarrow{a}=21$.

Answer 1

Hint: Here given that there are three vectors and we have to find the forth vector. Also given that the forth vector is perpendicular to all three vectors. When two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular to each other, then their dot product is zero, i.e., $\overrightarrow{a\cdot }\overrightarrow{b}=0$. By applying these properties we have to this question. For that first perform the dot-product of each vector and it will obtain the three equations.

Complete step by step solution:
Given, $\overrightarrow{a}=4\widehat{i}+5\widehat{j}-\widehat{k},\overrightarrow{b}=\widehat{i}-4\widehat{j}+5\widehat{k}$ and $\overrightarrow{c}=3\widehat{i}+\widehat{j}-\widehat{k}$.
Since, the vector $\overrightarrow{d}$ is perpendicular to both $\overrightarrow{c}$ and $\overrightarrow{b}$.
$\therefore \overrightarrow{d\cdot }\overrightarrow{b}=0$ and $\overrightarrow{d\cdot }\overrightarrow{c}=0$
Let $\overrightarrow{d}=x\widehat{i}+y\widehat{j}+z\widehat{k}$
Now, $\overrightarrow{d\cdot }\overrightarrow{b}=0$
$\Rightarrow \left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)\cdot \left( \widehat{i}-4\widehat{j}+5\widehat{k} \right)=0$
$\Rightarrow x-4y+5z=0\ldots .\text{ }\left( 1 \right)$
Similarly, $\overrightarrow{d\cdot }\overrightarrow{c}=0$
$\Rightarrow \left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)\cdot \left( 3\widehat{i}+\widehat{j}-\widehat{k} \right)=0$
$\Rightarrow 3x+y-z=0\ldots .\text{ }\left( 2 \right)$
And $\overrightarrow{d\cdot }\overrightarrow{a}=21$
$\Rightarrow \left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)\cdot \left( 4\widehat{i}+5\widehat{j}-\widehat{k} \right)=21$
\[\Rightarrow 4x+5y-z=21\ldots .\text{ }\left( 3 \right)\]
Now solve all the three equations,
$x-4y+5z=0\text{ }........\text{(1)}$
$3x+y-z=0\text{ }.........\text{(2)}$
$4x+5y-z=21\text{ }..........\text{(3)}$
Now, solve for ‘y’ variable in equation $\text{(2)}$
$\text{y = }-\text{3x + z }..........\text{(4)}$
Put this value of ‘y’ in equation \[\text{(3)}\], we get
$\Rightarrow$\[\text{4x + 5(}-3\text{x + z)}-\text{z = 21}\]
$\Rightarrow$\[\text{4x}-\text{15x +5z}-\text{z = 21}\]
$\Rightarrow$\[-11\text{x + 4z = 21 }..........\text{(5)}\]
Put the value of ‘y’ in equation \[\text{(1)}\] also, we get
$\Rightarrow$\[x-4\text{(}-3\text{x + z)}+5z=0\]
$\Rightarrow$\[x+12x-4z+5z=0\]
$\Rightarrow$\[13x+z=0\text{ }.............\text{(6)}\]
Solve equation \[\text{(6)}\] for ‘z’ variable, we get
$\Rightarrow$\[z=-13x\text{ }.........\text{(7)}\]
Put the value of ‘z’ in equation \[\text{(5)}\], we get
$\Rightarrow$\[-11x+4(-13x)=21\]
$\Rightarrow$\[-11x-52x=21\]
$\Rightarrow$\[-63x=21\]
$\Rightarrow$\[x=-\dfrac{21}{63}\]
$\Rightarrow$\[x=-\dfrac{1}{3}\]
Put the value of ‘x’ in equation \[\text{(7)}\], we get
\[z=-13x\]
$\Rightarrow$\[z=(-13)\times \left( -\dfrac{1}{3} \right)\]
$\Rightarrow$\[z=\dfrac{13}{3}\text{ }\]
Now, put the value of ‘x’ and ‘z’ in equation \[\text{(4)}\], we get
\[\text{y = }-\text{3x + z}\]
$\Rightarrow$\[y=-3\times \left( -\dfrac{1}{3} \right)+\dfrac{13}{3}\]
$\Rightarrow$\[y=1+\dfrac{13}{3}\]
$\Rightarrow$\[y=\dfrac{3+13}{3}\]
$\Rightarrow$\[y=\dfrac{16}{3}\]
So, on solving equation (1), (2) and (3), we get
$x=\dfrac{-1}{3},y=\dfrac{16}{3}$ and $z=\dfrac{13}{3}$.

So, $\overrightarrow{d}=x\widehat{i}+y\widehat{j}+z\widehat{k}=\dfrac{-1}{3}\widehat{i}+\dfrac{16}{3}\widehat{j}+\dfrac{13}{3}\widehat{k}=\dfrac{-\widehat{i}+16\widehat{j}+13\widehat{k}}{3}=\dfrac{1}{3}\left( -\widehat{i}+16\widehat{j}+13\widehat{k} \right)$

Note: Algebraically, the dot product of two vectors is the sum of the products of the corresponding entries of the two sequences of numbers. To find whether the given vector is perpendicular to another vector or not, we can adapt two methods that is, dot-product and cross-product of the vectors. The dot-product of vectors can be given as, \[\text{A = (}{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}{{\text{a}}_{\text{3}}}\text{) and B = (}{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}{{\text{b}}_{\text{3}}}\text{)}\]it is equal to the sum of product of corresponding numbers or terms which is,\[\text{A}\centerdot \text{B = }{{\text{a}}_{\text{1}}}{{\text{b}}_{\text{1}}}\text{ + }{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}\text{ + }{{\text{a}}_{\text{3}}}{{\text{b}}_{\text{3}}}\]. If the dot-product of two vectors is equal to zero then that vectors are perpendicular vectors. Whereas, the cross-product can be done as,\[\text{A}\times \text{B = (}{{\text{a}}_{2}}{{\text{b}}_{3}}-\text{ }{{\text{a}}_{3}}{{\text{b}}_{\text{2}}}\text{, }{{\text{a}}_{\text{3}}}{{\text{b}}_{1}}-{{\text{a}}_{1}}{{\text{b}}_{3}}\text{, }{{\text{a}}_{1}}{{\text{b}}_{2}}-{{\text{a}}_{2}}{{\text{b}}_{1}}).\]