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
$\begin{pmatrix}0 & -c_3 & c_2 \\c_3 & 0 & -c_1 \\-c_2 & c_1 & 0\end{pmatrix}\begin{pmatrix} 1 \\b_2 \\b_3\end{pmatrix}=\begin{pmatrix}3-c_1 \\1-c_2 \\-1-c_3\end{pmatrix} $.
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. $|\vec{c}| \leq \sqrt{11}$

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 } $
$ \begin{pmatrix} 0 & -c_3 & c_2 \\c_3 & 0 & -c_1 \\-c_2 & c_1 & 0\end{pmatrix}\begin{pmatrix} 1 \\b_2 \\b_3\end{pmatrix}=\begin{pmatrix} 3-c_1 \\1-c_2 \\-1-c_3\end{pmatrix} $
multiply & compare
$b_2 c_3-b_3 c_2=c_1-3 .....$(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 } \neq 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}| \leq \sqrt{11} $
given $ \vec{a} \cdot \vec{b}=0$
$ b _2- b _3=-3 $ also
$ b _2^2+ b _3^2-2 b _2 b _3=9 \,\, b _2 b _3>0$
$ b _2^2+ b _3^2=9+2 b _2 b _3$
$ b _2^2+ b _3^2=9+2 b _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}$