Question

If $a$ and $b$ are two vectors such that$\left|a+b\right|=\sqrt{29}$ and$a\times (2i+3j+4k)=(2i+3j+4k)\times b$, a possible value of $(a+b).(-7i+2j+3k)$ is

• A. $0$
• B. $3$
• C. $4$
• D. $8$

Answer 1

The correct option is C

$4$

Explanation for the correct option:

Step 1: Find the vector$\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)}$.

We have given$\left|a+b\right|=\sqrt{29}$ and $a\times (2i+3j+4k)=(2i+3j+4k)\times b$,

we have to find a possible value of$(a+b)·(-7i+2j+3k)$

now,

$\begin{array}{l}a\times (2i+3j+4k)=(2i+3j+4k)\times b\end{array}$

$\Rightarrow$ $a\times (2i+3j+4k)=-b\times (2i+3j+4k)$ $\left[\because a\times b=-b\times a\right]$

$\Rightarrow$$(a+b)\times (2i+3j+4k)=0$

$\therefore (a+b)\parallel (2i+3j+4k)$

Step 2: Find $\mathit{\lambda }\mathbf{\&}\mathbf{}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)}$.

$\begin{array}{rcl}(a+b)& =& \lambda (2i+3j+4k)\end{array}$

$\Rightarrow$$\begin{array}{rcl}\left|a+b\right|& =& \left|\lambda (2i+3j+4k)\right|\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\left(\sqrt{29}\right)& =& \lambda \left(\sqrt{4+9+16}\right)\end{array}$

$\Rightarrow$$\begin{array}{rcl}{\left(\sqrt{29}\right)}^2& =& {\lambda }^2{\left(\sqrt{29}\right)}^2\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{29}{29}& =& {\lambda }^2\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\lambda & =& \pm 1\end{array}$

$\Rightarrow$ $(a+b)=\pm 1(2i+3j+4k)$

Step 3: Find the value of$\mathbf{}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)}\mathbf{·}\mathbf{(}\mathbf{-}\mathbf{7}\mathit{i}\mathbf{+}\mathbf{2}\mathit{j}\mathbf{+}\mathbf{3}\mathit{k}\mathbf{)}$.

$\begin{array}{rcl}\therefore (a+b).(-7i+2j+3k)& =& \pm (2i+3j+4k).(-7i+2j+3k)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(a+b).(-7i+2j+3k)& =& \pm (-14+6+12)\end{array}$

$\Rightarrow$$\begin{array}{rcl}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{.}\mathbf{(}\mathbf{-}\mathbf{7}\mathbf{i}\mathbf{+}\mathbf{2}\mathbf{j}\mathbf{+}\mathbf{3}\mathbf{k}\mathbf{)}& \mathbf{=}& \mathbf{\pm }\mathbf{4}\end{array}$

Hence, the correct option is (C)