Question

If $x\times a+kx=b$ where $k$ is a scalar and $a,b$ are any two vectors, then determine $x$ in terms of $a,b$ and $k.$

• A. $x=\frac{1}{(a^2+k^2)}[kb-(a\times b)+\frac{(a.b)}{k}a]$
• B. $x=\frac{1}{(a^2+k^2)}[ka-(a\times b)+\frac{(a.b)}{k}b]$
• C. $x=\frac{1}{(a^2+k^2)}[ka+(a\times b)+\frac{(a.b)}{k}b]$
• D. $x=\frac{1}{(a^2+k^2)}[kb+(a\times b)+\frac{(a.b)}{k}a]$

Answer 1

The correct option is D $x=\frac{1}{(a^2+k^2)}[kb+(a\times b)+\frac{(a.b)}{k}a]$

$x\times a+kx=b$.....(1)
Pre-multiply the given equation vectorially by $a$

$a\times (x\times a)+k(a\times x)=a\times b$ $\Rightarrow (a.a)x-(a.x)a+k(a\times x)=a\times b$ .....(2)
Again pre-multiply the given equation scalarly by $a$

$[a\times x]+k(a.x)=a.b$ $\Rightarrow 0+k(a.x)=a.b$ ......(3)
In order to eliminate the scalar $(a.x)$ between (2) and (3),

we multiply (2) by $k$ and (3) by $a$ and add
$a^{2}kx+k^2(a\times x)=k(a\times b)+(a.b)a$
$a^{2}kx+k^2(kx-b)=k(a\times b)+(a.b)a$ ....(4)
But from (1), $kx-b=-(x\times a)=(a\times x)$
Putting the value of $a\times x$ in (4),

we get $a^{2}kx+k^2(kx-b)=k(a\times b)+(a.b)a$
$\Rightarrow k(a^2+k^2)x=k^{2}b+k(a\times b)+(a.b)a$
$\therefore x=\frac{1}{(a^2+k^2)}[kb+(a\times b)+\frac{(a.b)}{k}a]$