Question

The vector perpendicular to the vectors $4\hat{i}-\hat{j}+3\hat{k}$ and $-2\hat{i}+\hat{j}-2\hat{k}$ whose magnitude is $9$

• A. $3\hat{i}+6\hat{j}-6\hat{k}$
• B. $3\hat{i}-6\hat{j}+6\hat{k}$
• C. $-3\hat{i}+6\hat{j}+6\hat{k}$
• D. None of the above

Answer 1

The correct option is C $-3\hat{i}+6\hat{j}+6\hat{k}$

Let

$a=4\hat{i}-\hat{j}+3\hat{k},$

$b=-2\hat{i}+\hat{j}-2\hat{k}$ and

$c=x\hat{i}+y\hat{j}+z\hat{k}$

Given,

$a\cdot c=0$
i.e., $4x-y+3x=0$ ......(i)
and $b\cdot c=0$
i.e. $-2x+y-2z=0$ ......(ii)

Also, $|c|=9$
i.e. $x^2+y^2+z^2=81$ .......(iii)

Now, from Eqs. (i) and (ii), we get
$2x+z=0\Rightarrow z=-2x$

On putting this value in Eq. (iii), we get
$x^2+y^2+4x^2=81$
$\Rightarrow 5x^2+y^2=81$ ......(iv)

On multiplying Eq. (i) by $2$ and Eq. (ii) by $3$ and then adding, we get
$8x-3y+6z=0$
$\underset{–}{-6x+3y-6z=0}$
$2x+y=0$
$\Rightarrow y=-2x$

On putting this value in Eq. (iv), we get
$5x^2+4x^2=81$
$\Rightarrow 9x^2=81$
$\Rightarrow x^2=9$
$\Rightarrow x=\pm 3$
$\therefore y=\mp 6$ and $z=\mp 6$

$\therefore$ Required vector, $c=xi+yj+zk$
$=\pm 3\hat{i}\mp 6\hat{j}\mp 6\hat{k}$
$=3\hat{i}-6\hat{j}-6\hat{k}$
or
$=-3\hat{i}+6\hat{j}+6\hat{k}$