Question

what is the value of n so that vectors 2i+3j-2k ,5i+nj+k ,-i+2j+3 k may be colinear ​

Answer 1

Given:

The vectors 2i+3j-2k ,5i+nj+k ,-i+2j+3 k may be colinear ​

To find:

What is the value of n so that vectors 2i+3j-2k ,5i+nj+k ,-i+2j+3 k may be colinear ​

Solution:

If  a, b, c   are coplanar, then their scalar triple product should be zero.

[a b c ] = 0  

That implies,

(a × b ) . c = 0  

By cyclic rotation,

(c × a) . b = 0

From given, we have,

2i + 3j - 2k , 5i + nj + k , -i + 2j + 3

Let a = 2i + 3j - 2k

b = 5i + nj + k

c = -i + 2j + 3

Now consider,

(c × a) = (-i + 2j + 3) × (2i + 3j - 2k)

(c × a) = (13i + 8j - k)

Now consider,

(c × a) . b = 0

(13i + 8j - k) . (5i + nj + k) = 0

upon solving, we get,

65 + 8n - 1 = 0

64 + 8n = 0

n = -64/8

n = - 8

Therefore, the value of n so that vectors 2i+3j-2k ,5i+nj+k ,-i+2j+3 k may be colinear ​is -8.