The correct option is
C 115(−11i−10j−2k)Let
^c=x^i+y^j+z^kwhere x2+y2+z2=1......(1)
And unit vector along 3^i+4^j=3^i+4^j√32+42=3^i+4^j5
The bisectors of these two is given by
r=t(^a+^b)
r=t(x^i+y^j+x^k+3^i+4^j5)
r=t5[(5x+3)^i+(5y+4)^j+5z^k).......(2)
But the bisector is given by −^i+^j−^k.......(3)
Comparing (2) and (3), we get
t5(5x+3)=−1⇒x=−5+3t5t
t5(5y+4)=1⇒y=5−4t5t
t5(5z)=−1⇒z=−1t
Put all the values in equation (1), we get
(−5+3t5t)2+(5−4t5t)2+(−1t)2=1
25+9t2+30t+25+16t2−40t+2525t2=1
25t2−10t+75=25t2
t=7.5
Thus,
x=−5+3×7.55×7.5=−1115
y=5−4×7.55×7.5=−1015
z=−17.5=−215
Hence, ^c=115(−11^i−10^j−2^k)