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CBSE Class 12 Math 2011 Solved Paper

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Question : 20 of 29
Marks: +1, -0
Find a unit vector perpendicular to each of the vector
→
a
+
→
b
 and
→
a
−
→
b
 , where
→
a
 = 3
^
i
+2
^
j
+2
^
k
 and
→
b
 =
^
i
+2
^
j
−2
^
k
Solution:  
→
a
 = 3
^
i
+2
^
j
+2
^
k
 ,
→
b
 =
^
i
+2
^
j
−2
^
k
∴
→
a
+
→
b
 = 4
^
i
+4
^
j
 and
→
a
−
→
b
 = 2
^
i
+4
^
k
(
→
a
+
→
b
)×(
→
a
−
→
b
) = |
^
i
^
j
^
k
440
204
|
=
^
i
 (16) -
^
j
 (16) +
^
k
 (-8) = 16
^
i
−16
^
j
−8
^
k
∴ |(
→
a
+
→
b
)×(
→
a
−
→
b
)| = √162+(−16)2+(−8)2
= √256+256+64
= √576 = 24
So the unit vector, perpendicular to each of the vectors
→
a
+
→
b
 and
→
a
−
→
b
 is given by ±
(
→
a
+
→
b
)×(
→
a
−
→
b
)
|(
→
a
+
→
b
)×(
→
a
−
→
b
)|
 = ±
16
^
i
−16
^
j
−8
^
k
24
 = ±
2
^
i
−2
^
j
−
^
k
3
 = ±
2
3
^
i
∓
2
3
^
j
∓
1
3
^
k
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