Question

If the vector $-i+j-k$ bisects the angle between the vectors c and the vector $3i+4j$ then find the unit vector along c.

• A. $-\frac{1}{15}(-11i+10j+2k)$
• B. $-\frac{1}{15}(11i-10j+2k)$
• C. $-\frac{1}{15}(-11i+10j-2k)$
• D. $-\frac{1}{15}(11i+10j+2k)$

Answer 1

The correct option is D $-\frac{1}{15}(11i+10j+2k)$

Let the unit vector along c be

$xi+yj+zk,$ where $\sum x^2=1$

$\therefore -i+j-k=t[xi+yj+zk+\frac{3i+4j}{5}]$

Comparing coefficients,

$-1=t(x+\frac{3}{5}),1=t(y+\frac{4}{5}),-1=tz$

Since

$x^2+y^2+z^2=1,$

$wehavei.e.$

$(-\frac{1}{t})-{\left(\frac{3}{5}\right)}^2$$(-\frac{1}{t})-{\left(\frac{4}{5}\right)}^2$+ ${(-\frac{1}{t})}^2=1$

or $\frac{3}{t^2}+\frac{2}{t}(\frac{3}{5}-\frac{4}{5})+\frac{9+16}{25}=1$or $15-2t=0\therefore t=\frac{15}{2}$

$\therefore x=-\frac{2}{15}-\frac{3}{5}=-\frac{11}{15},y=\frac{2}{15}-\frac{4}{5}=-\frac{10}{15},z=-\frac{2}{15}$

$\therefore c=xi+yj+zk=-\frac{1}{15}(11i+10j+2k)$