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Question

The vector i+j+k bisects the angle between the vectors ¯¯c and 3i+4j. Then unit vector in the direction of ¯¯c is ___

A
115(11i+10j2k)
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B
115(11i10j+2k)
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C
115(11i10j2k)
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D
115(11i+10j+2k)
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Solution

The correct option is C 115(11i10j2k)
Let ^c=x^i+y^j+z^k
where x2+y2+z2=1......(1)
And unit vector along 3^i+4^j=3^i+4^j32+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)=1x=5+3t5t
t5(5y+4)=1y=54t5t
t5(5z)=1z=1t
Put all the values in equation (1), we get
(5+3t5t)2+(54t5t)2+(1t)2=1
25+9t2+30t+25+16t240t+2525t2=1
25t210t+75=25t2
t=7.5
Thus,
x=5+3×7.55×7.5=1115
y=54×7.55×7.5=1015
z=17.5=215
Hence, ^c=115(11^i10^j2^k)



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