If a^{\to} = i^{\^}+j^{\^}+k^{\^} and b^{\to} = j^{\^}-k^{\^} , find a vector c^{\to} such that a^{\to}\timesc^{\to} = b^{\to} and a^{\to}.c^{\to} = 3 OR If a^{\to}+b^{\to}+c^{\to} = 0 and |a^{\to}| = 3 , |b^{\to}| = 5 and |c^{\to}| = 7, show that the angle between a^{\to} and b^{\to} is 60°

CBSE Class 12 Math 2008 Solved Paper - Question 20

Question : 20 of 29

Marks: +1, -0

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↖{→}

xi↖{\^}+yj↖{\^}+zk↖{\^}

a↖{→}

i↖{\^}+j↖{\^}+k↖{\^}

∴

a↖{→}×c↖{→}

|\table i↖{\^},j↖{\^},k↖{\^};1,1,1;x,y,z|

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↖{\^}

xi↖{\^}+yj↖{\^}+zk↖{\^}

= 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↖{→}

5/3i↖{\^}+2/3j↖{\^}+2/3k↖{\^}

Thus, the required vector

c↖{→}

5/3i↖{\^}+2/3j↖{\^}+2/3k↖{\^}

a↖{→}+b↖{→}+c↖{→}

= 0 ⇒

a↖{→}+b↖{→}

-c↖{→}

a↖{→}+b↖{→}.a↖{→}+b↖{→}

-c↖{→}.-c↖{→}

a↖{→}.a↖{→}

2a↖{→}.b↖{→}

b↖{→}.b↖{→}

c↖{→}.c↖{→}

|a↖{→}|^2 + 2 |a↖{→}||b↖{→}|

cos θ +

|b↖{→}|^2

|c↖{→}|^2

3^2+(2)(3)(5) cosθ + 5^2

7^2

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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