Question

If a vector 'r' of magnitude $3\sqrt{6}$ is directed along the bisector of the angle between the vectors a = 7i - 4j - 4 k and b = -2i -j + 2k, then r is equal to

• A. i - 7j + 2k
• B. i + 7j - 2k
• C. i + 7j + 2k
• D. i - 7j - 2k

Answer 1

The correct option is A i - 7j + 2k

$\frac{7x-4y-4k}{\sqrt{81}}=1\left[\frac{-2x-y+2z}{\sqrt{3}}\right]$

$\frac{7x-4y-4k}{9}=1\left[\frac{-2x-4+2z}{3}\right]$

$7x-4y-4k=\pm [-6x-3y+6z]$

takes the

$7x-4y-4k=-6x-3y+6z$

$13x-y-10k=0$

take ${-}^{\prime }v{e}^{\prime }$

$7x-4y-4k=6x+3y-6z$

$x-7y+2z=0$

magnitude $=\sqrt{1+49+4}=\sqrt{54}=2\sqrt{13}$

So it should be it is Answer