Question

Let $\hat{i},\hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let
$\overrightarrow{a}=3\hat{i}+\hat{j}-\hat{k},$
$\overrightarrow{b}=\hat{i}+b_2\hat{j}+b_3\hat{k},b_2,b_3\in R,$
$\overrightarrow{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k},c_1,c_2,c_3\in R$
be three vectors such that $b_2{b}_3>0,\vec{a}\cdot \vec{b}=0$ and
$\left(\begin{array}{ccc}0& -c_3& c_2\\ c_3& 0& -c_1\\ -c_2& c_1& 0\end{array}\right)\left(\begin{array}{c}1\\ b_2\\ b_3\end{array}\right)=\left(\begin{array}{c}3-c_1\\ 1-c_2\\ -1-c_3\end{array}\right)$.
Then, which of the following is/are TRUE?

• A. $\vec{a}\cdot \vec{c}=0$
• B. $\overrightarrow{b}\cdot \overrightarrow{c}=0$
• C. $|\vec{b}|>\sqrt{10}$
• D. $|\vec{c}|\le \sqrt{11}$

Answer 1

Answer: (B), (C), (D)

$\vec{a}=3\hat{i}+\hat{j}-\hat{k}$
$\vec{b}=\hat{i}+b_2\hat{j}+b_3\hat{k}$
$\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$
$\left(\begin{array}{ccc}0& -c_3& c_2\\ c_3& 0& -c_1\\ -c_2& c_1& 0\end{array}\right)\left(\begin{array}{c}1\\ b_2\\ b_3\end{array}\right)=\left(\begin{array}{c}3-c_1\\ 1-c_2\\ -1-c_3\end{array}\right)$
multiply & compare
$b_2{c}_3-b_3{c}_2=c_1-\mathrm{3.....}$(1)
$c_3-b_3{c}_1=1-c_2.......$(2)
$c_2-b_2{c}_1=1+c_3......$(3)
(1) $\hat{i}-(2)\hat{j}+(3)\hat{k}$
$\hat{i}\left(b_2{c}_3-b_3{c}_2\right)-\hat{j}\left(c_3-b_3{c}_1\right)+\hat{k}\left(c_2-b_2{c}_1\right)$
$=c_1\hat{i}+c_2\hat{j}+c_2\hat{k}-3\hat{i}-\hat{j}+\hat{k}$
$\vec{b}\times \vec{c}=\vec{c}-\vec{a}$
Take dot product with $\vec{b}$
$0=\vec{c}\cdot \vec{b}-\vec{a}\cdot \vec{b}$
$\vec{b}\cdot \vec{c}=0$
$\vec{b}\perp \vec{c}$
${\vec{b}}^{\wedge }\vec{c}=90^{\circ }$
Take dot product with $\vec{c}$
$0=|\vec{c}{|}^2-\vec{a}\cdot \vec{c}$
$\vec{a}\cdot \vec{c}=|\vec{c}{|}^2$
$\vec{a}\cdot \vec{c}\ne 0$
$\vec{b}\times \vec{c}=\vec{c}-\vec{a}$
Squaring
$|\vec{b}{|}^2|\vec{c}{|}^2=|\vec{c}{|}^2+|\vec{a}{|}^2-2\vec{c}\cdot \vec{a}$
$|\vec{b}{|}^2|\vec{c}{|}^2=|\vec{c}{|}^2+11-2|\vec{c}{|}^2$
$|\vec{b}{|}^2|\vec{c}{|}^2=11-|\vec{c}{|}^2$
$|\vec{c}{|}^2\left(|\vec{b}{|}^2+1\right)=11$
$|\vec{c}{|}^2=\frac{11}{|\vec{b}{|}^2+1}$
$|\vec{c}|\le \sqrt{11}$
given $\vec{a}\cdot \vec{b}=0$
$b_2-b_3=-3$ also
$b_2^2+b_3^2-2b_2{b}_3=9b_2{b}_3>0$
$b_2^2+b_3^2=9+2b_2{b}_3$
$b_2^2+b_3^2=9+2b_2{b}_3>9$
$b_2^2+b_3^2>9$
$|\vec{b}|=\sqrt{1+b_2^2+b_3^2}$
$|\vec{b}|>\sqrt{10}$