Question

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

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

Answer 1

Answer: (C)

Let $a=4i-j+3k,b=-2i+j-2k$
and $c=xi+yj+zk$
Given, $a\cdot c=0$
i.e., $4x-y+3z=0\dots (i)$
and $b\cdot c=0$
i.e., $-2x+y-2z=0\dots (ii)$
Also, $|c|=9$
i.e., $x^2+y^2+z^2=81\dots (iii)$
Now, from Eqs. (i) and (ii), we get
$2x+z=0$
$\Rightarrow z=-2x$
On putting this value in Eq. (iii), we get
$x^2+y^2+4x^2=81$
$\Rightarrow 5x^2+y^2=81\dots (iv)$
On multiplying Eq. (i) by 2 and Eq. (ii) by 3 and then adding, we get
$8x-2y+6z=0$
$-6x+3y-6z=0$
______________
$2x+y=0$
$\Rightarrow y=-2x$
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, $c=xi+yj+zk$
$=\pm 3i\mp 6j\mp 6k$
$=3i-6j-6k$
or
$=-3i+6j+6k$