Question

If the solution of the vector equation $r\times a+kr=b$, where $a$ and $b$ are two given vectors and $k$ is a scalar quantity, is represented by $r=\frac{1}{k^2+|a{|}^2}[\frac{(a\cdot b)a}{k}+(p)b+(a\times b\left)\right]$, then $p$ is equal to

• A. $(a\cdot b)$
• B. $k^2$
• C. $k$
• D. $k(a\cdot b)$

Answer 1

The correct option is C $k$

If $\overline{a},\overline{b}$ are linearly independent

Then, vector $\overrightarrow{r}=x\overrightarrow{a}+y\overline{b}+3(\overline{a}\times \overline{b})$

Given, $\overline{r}\times \overline{a}+kr=b$

$\Rightarrow (x\overline{a}+y\overline{b}+3(\overline{a}\times \overline{b}\left)\right)\times \overline{a}+k(x\overline{a}+y\overline{b}+3(\overline{a}\times b\left)\right)=\overline{b}$

$\Rightarrow y(\overline{b}\times \overline{a})+3({\left|a\right|}^2\overline{b}-(\overline{a}.\overline{b}\left)\right)\overline{a}+kx\overline{a}$ $+ky\overline{b}+k3(\overline{a}\times \overline{b})=\overline{b}$

$\Rightarrow (kx-3(\overline{a}.\overline{b}\left)\right)\overline{a}+(3{\left|a\right|}^2+ky)\overline{b}$ $+(k3-y)(\overline{a}\times \overline{b})=\overline{b}$

Comparing the co-efficeints

$\Rightarrow kx=3(\overline{a}.\overline{b}),$ $k3=y,$ $3a^2+ky=1$

$\Rightarrow 3a^2+k\times k3=1$

$\Rightarrow 3(a^2+k^2)=1$

$\Rightarrow 3=\frac{1}{a^2+k^2}$

$⟹y=k/a^2+k^2$

$\therefore p=k$