Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=x_1 \hat{i}+x_2 \hat{j}+x_3 \hat{k}$, where $x_1, x_2,

Question

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=x_1 \hat{i}+x_2 \hat{j}+x_3 \hat{k}$, where $x_1, x_2, x_3 \in\{-3,-2,-1,0,1,2\}$. Number of possible vectors $\vec{b}$ such that $\vec{a}$ and $\vec{b}$ are mutually perpendicular, is p then $\frac{p}{5}=$

Professor PC Thomas Classes · Answer

$
\vec{a} \cdot \vec{b}=0 x_1+x_2+x_3=0
$
Thus, we have to obtain the number of integral solutions of this equation.
Coefficient of $x^0$ in $\left(x^{-3}+x^{-2}+x^{-1}+x^0+x+x^2\right)^3$
$=x^0\left|\left(\frac{1+x+x^2+x^3+x^4+x^5}{x^3}\right)^3=x^9\right|$
$\left(1-x^6\right)^3(1-x)^{-3}={ }^{11} C_9-3 \cdot{ }^5 C_3=25$.