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CBSE Class 12 Math 2008 Solved Paper

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Question : 20
Total: 29
If
→
a
 =
^
i
+
^
j
+
^
k
 and
→
b
 =
^
j
−
^
k
 , find a vector
→
c
 such that
→
a
×
→
c
 =
→
b
 and
→
a
.
→
c
 = 3
OR
If
→
a
+
→
b
+
→
c
 = 0 and |
→
a
| = 3 , |
→
b
| = 5 and |
→
c
| = 7, show that the angle between
→
a
 and
→
b
 is 60°
Solution:  
Let
→
c
 = x
^
i
+y
^
j
+z
^
k
→
a
 =
^
i
+
^
j
+
^
k
∴
→
a
×
→
c
 = |
^
i
^
j
^
k
111
xyz
|
→
a
×
→
c
 =
^
i
(z−y)−
^
j
(z−x)+
^
k
(y−x) ... (1)
Now,
→
a
×
→
c
 =
→
b
→
b
 =
^
j
−
^
k
 ... (2)
Comparing (1) and (2), we get :
z – y = 0 ⇒ z = y ...(3)
z – x = -1 ...(4)
y – x = -1 ...(5)
Also, given that
→
a
.
→
c
 = 3
∴
^
i
+
^
j
+
^
k
 . x
^
i
+y
^
j
+z
^
k
 = 3
x + y + z = 3
Using (3), we get, x + 2y = 3 ...(6)
Adding (5) and (6), we get
3y = 2 ⇒ y =
2
3
∴ z =
2
3
 Since z = y
From (6), we have,x = 3 - 2y
⇒ x = 3 -
2×2
3
⇒ x =
9−4
3
⇒ x =
5
3
∴
→
c
 =
5
3
^
i
+
2
3
^
j
+
2
3
^
k
Thus, the required vector
→
c
 is
5
3
^
i
+
2
3
^
j
+
2
3
^
k
OR
→
a
+
→
b
+
→
c
 = 0 ⇒
→
a
+
→
b
 = −
→
c
→
a
+
→
b
.
→
a
+
→
b
 = −
→
c
.−
→
c
→
a
.
→
a
 + 2
→
a
.
→
b
 +
→
b
.
→
b
 =
→
c
.
→
c
|
→
a
|2+2|
→
a
||
→
b
| cos θ + |
→
b
|2 = |
→
c
|2
32+(2)(3)(5)cosθ+52 = 72
9 + 30cos θ + 25 = 49
30 cos θ = 15 ⇒ cos θ =
1
2
cos θ = cos 60° ⇒ θ = 60°
Hence proved.
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