Question

If $\overrightarrow{a}=i-j,\overrightarrow{b}=i+j+k$ are two vectors and $c$ is another vector such that $\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}=\overrightarrow{0}$and $\overrightarrow{a}.\overrightarrow{c}=4$, then ${\overrightarrow{c}}^2=$

• A. $9$
• B. $8$
• C. $\frac{19}{2}$
• D. $\frac{17}{2}$

Answer 1

The correct option is C

$\frac{19}{2}$

Explanation for the correct option:

Step 1: Take cross-product $\overrightarrow{a}\times \overrightarrow{c}=-\overrightarrow{b}$both sides with $\overrightarrow{a}$.

We have given $\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}=\overrightarrow{0}$ and $\overrightarrow{a}.\overrightarrow{c}=4$

we have to find:${\left|\overrightarrow{c}\right|}^2$

now, $\overrightarrow{a}\times \overrightarrow{c}=-\overrightarrow{b}$

taking cross-product both sides with $\overrightarrow{a}$.

$\begin{array}{rcl}\overrightarrow{(a}\times \overrightarrow{c)}\times \overrightarrow{a}& =& -\overrightarrow{b}\times \overrightarrow{a}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\overrightarrow{a}\mathbf{}\times (\overrightarrow{a}\times \overrightarrow{c})& =& \overrightarrow{b}\times \overrightarrow{a}\end{array}$ $\begin{array}{rcl}& & \left[\mathbf{\because }\mathbf{}\mathbf{(}\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\overrightarrow{\mathbf{b}}\mathbf{\times }\overrightarrow{\mathbf{a}}\right]\end{array}$

$\Rightarrow$$\begin{array}{rcl}(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{a}-(\overrightarrow{a}.\overrightarrow{a})\overrightarrow{c}& =& \overrightarrow{b}\times \overrightarrow{a}\end{array}$ $\begin{array}{rcl}& & \left[\mathbf{\because }\overrightarrow{\mathbf{a}}\mathbf{}\mathbf{\times }\mathbf{(}\overrightarrow{\mathbf{b}}\mathbf{\times }\overrightarrow{\mathbf{c}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{(}\overrightarrow{\mathbf{a}}\mathbf{.}\overrightarrow{\mathbf{c}}\mathbf{)}\overrightarrow{\mathbf{b}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{(}\overrightarrow{\mathbf{a}}\mathbf{.}\overrightarrow{\mathbf{b}}\mathbf{)}\overrightarrow{\mathbf{c}}\right]\end{array}$

$\Rightarrow$ $\begin{array}{rcl}4\overrightarrow{a}-2\overrightarrow{c}& =& \overrightarrow{b}\times \overrightarrow{a}.....\left(1\right)\end{array}$

Step 2: Take the dot product of$\overrightarrow{a}$ with itself.

$\begin{array}{rcl}\overrightarrow{a}.\overrightarrow{a}& =& (i-j)(i-j)\\ & =& 1+1\\ & =& 2\end{array}$

Step 3: take the cross product$\begin{array}{rcl}& & \overrightarrow{b}\times \overrightarrow{a}\end{array}$.

$\begin{array}{rcl}\overrightarrow{b}\times \overrightarrow{a}& =& \left|\begin{array}{ccc}i& j& k\\ 1& 1& 1\\ 1& -1& 0\end{array}\right|\\ & =& (0-(-1\left)\right)i-(0-1)j+(-1-1)k\end{array}$

$\Rightarrow$$\begin{array}{rcl}\overrightarrow{b}\times \overrightarrow{a}& =& i+j-2k......\left(2\right)\end{array}$

Step 4: take the dot product of$-2\overrightarrow{c}$ with itself.

From $\left(1\right)\&\left(2\right)$

$\begin{array}{l}4\overrightarrow{a}-2\overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{a}\end{array}$

$\Rightarrow$$\begin{array}{lll}4(i-j)-2\overrightarrow{c}& =& (i+j-2k)\end{array}$

$\Rightarrow$ $\begin{array}{lll}-2\overrightarrow{c}& =& -3i+5j-2k\end{array}$

Step 5. Taking dot-product with the same vector,

$\begin{array}{rcl}(-2\overrightarrow{c}).(-2\overrightarrow{c})& =& (-3i+5j-2k).(-3i+5j-2k)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}4{\left|\overrightarrow{c}\right|}^2& =& 9+25+4\end{array}$

$\begin{array}{rcl}\therefore {\left|\overrightarrow{c}\right|}^2& =& \frac{38}{4}\\ & =& \frac{\mathbf{19}}{\mathbf{2}}\end{array}$

hence, the option$\left(C\right)$ is correct.