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. \(\vec{a}.\vec{c}=0\)
• B. \(\vec{b}.\vec{c}=0\)
• C. \(\left | \vec{b} \right |> \sqrt10\)
• D. \(\left | \vec{c} \right |\leq  \sqrt11\)

Answer 1

The Correct Option is D

The correct answer is option (D): \(\left | \vec{c} \right |\leq  \sqrt11\)

Answer 2

Given,
\(\begin{pmatrix}0 & -c_3 & c_2 \\c_3 & 0 & -c_1 \\-c_2 & c_1 & 0\end{pmatrix}\begin{pmatrix} 1 \\b_2 \\b_3\end{pmatrix}=\begin{pmatrix}3-c_1 \\1-c_2 \\-1-c_3 \end{pmatrix}\)

\(\overrightarrow{ a }=3 \hat{ i }+\hat{ j }-\hat{ k }\\ \overrightarrow{ b }=\hat{ i }+ b _2 \hat{ j }+ b _3 \hat{ k }\\ \overrightarrow{ c }= c _1 \hat{ i }+ c _2 \hat{ j }+ c _3 \hat{ k }\)

Solve the Matrix
b₂​c₃​−b₃​c₂​ = c₁​−3 .....(1)
c₃​−b₃​c_1​ = 1−c₂​ .......(2)
c₂​−b₂​c₁​ = 1+c₃​ ......(3)

The vectors \(\overrightarrow{a}\),\(\overrightarrow{b}\), and \(\overrightarrow{c}\) are represented in terms of their components along the \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) axes respectively.
\((1) \ \hat{i} - (2) \hat{j} + (3) \hat{k}\)

\(\hat{i} (b_2c_3 - b_3c_2) - \hat{j} (c_3 - b_3c_1) + \hat{k} (c_2 - b_2c_1)\)
\(= c_1 \hat{i} + c_2 \hat{j} + c_2 \hat{k} - 3 \hat{i} - \hat{j} + \hat{k}\)
\(\vec{b} \times \vec{c} = \vec{c} - \vec{a}\)
Equating to \(\overrightarrow{c} - \overrightarrow{a}\):

Take dot product with \(\vec{b}\)
\(0 = \vec{c} \cdot \vec{b} - \vec{a} \cdot \vec{b}\)
\(\vec{b} \cdot \vec{c} = 0\)
\(\vec{b} \perp \vec{c}\)
\(\vec{b} \land \vec{c} = 90^\circ\)

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 the equation:
\(\begin{aligned} & |\vec{b}|^2|\vec{c}|^2=|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a} \\ & |\vec{b}|^2|\vec{c}|^2=|\vec{c}|^2+11-2|\vec{c}|^2 \\ & |\vec{c}|^2\left(|\vec{b}|^2+1\right)=11 \\ & |\vec{c}|^2=\frac{11}{|\vec{b}|^2+1} \\ & |\vec{c}|{\leq \sqrt{11}} \end{aligned}\)

Using \(\vec{a} \cdot \vec{b}=0\)
\(\begin{aligned} & b_2-b_3=-3 \text { also } \\ & b_2^2+b_3^2-2 b_2 b_3=9 b_2 b_3>0 \\ & b_2^2+b_3^2=9+2 b_2 b_3 \\ & b_2^2+b_3^2=9+2 b_2 b_3>9 \\ & b_2^2+b_3^2>9 \\ & |\vec{b}|=\sqrt{1+b_2^2+b_3^2} \\ & |\vec{b}|>\sqrt{10} \end{aligned}\)

S, the correct option is (B):\(\vec{b}.\vec{c}=0\), (C):\(\left | \vec{b} \right |> \sqrt10\) and (D): \(\left | \vec{c} \right |\leq  \sqrt11\)