what is the angle between A vector=5 i hat - 5 j hat and B vector = 5 i hat - 5 j hat
Answers
Explanation:
Let
c
^
=x
i
^
+y
j
^
+z
k
^
. where x
2
+y
2
+z
2
=1......(2)
And unit vector along 3
i
^
+4
j
^
=
3
2
+4
4
3
i
^
+4
j
^
=
5
3
i
^
+4
j
^
The bisectors of these two is given by
r=t(
a
^
+
b
^
)
⇒r=t(x
i
^
+y
j
^
+z
k
^
+
5
3
i
^
+4
j
^
)
⇒r=
5
t
((5x+3)
i
^
+(5y+4)
j
^
+5
k
^
) ..... (2)
But the bisector is given by −
i
^
+
j
^
−
k
^
........ (3)
Comparing (2) and (3), we get,
5
t
(5x+3)=−1⇒x=−
5t
5+3t
5
t
(5y+4)=1⇒y=
5t
5−4t
5
t
=−1⇒z=−
t
1
Put all the value in equation (1), we have,
(−
5t
5+3t
)
2
+(
5t
5−4t
)
2
+(−
t
1
)
2
=1
⇒
25t
2
25+9t
2
+30t+25+16t
2
−40t+25
=1
⇒25t
2
−10t+75=25t
2
⇒t=7.5
Thus,
x=−
5t
5+3t
⇒x=−
5×7.5
5+3×7.5
=−
37.5
5+22.5
=−
15
11
y=
5t
5−4t
⇒y=
5×7.5
5−4×7.5
=
37.5
5−30
=−
15
10
z=−
t
1
⇒z=−
7.5
1
=−
15
2
hence,
c
^
=−
15
11
i
^
−
15
10
j
^
−
15
2
k
^
Both the vectors coincide with each other. Thus the angle between them is equal to 0°